NCERT Solutions for Class 11 Chemistry Chapter 6 Equilibrium

Question 6.1(a) A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

What is the initial effect of the change on vapour pressure?

Answer :

By increasing the volume of the container suddenly, initially, the vapour pressure would decrease. It is due to, the amount of vapour has remained the same but it is distributed in a larger volume.

Question 6.1(b) A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

How do rates of evaporation and condensation change initially?

Answer :

Here the temperature is constant so that the rate of evaporation is also the same as before. On increasing the volume of the container, the density of vapour decreases due to which the rate of collision between vapour particles decreases. Hence the condensation rate also decreases initially.

Question 6.1(c) A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased .

What happens when equilibrium is restored finally and what will be the final vapour pressure?

Answer :

When equilibrium is restored, the rate of evaporation is equal to the rate of condensation. The temperature is constant and the volume is changed. The vapour pressure is temperature-dependent, not volumes. So, that final vapour pressure is equal to the initial vapour pressure.

Question 6.2 What is K c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂ ]= 0.60M, [O₂ ] = 0.82M and [SO₃ ] = 1.90M ?

$2 \mathrm{SO}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{SO}_3(g)$

Answer :

Followings are the given information values to solve the above problems-$\left[\mathrm{SO}_2\right]=0.60 \mathrm{M}\left[\mathrm{O}_2\right]$

$=0.82 \mathrm{M}\left[\mathrm{SO}_3\right]=1.90 \mathrm{M}$
The given chemical equation is -

$2 \mathrm{SO}_2(g)+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(g)$

The equilibrium constant for this reaction is expressed as;
$K_c=\frac{\left[\mathrm{SO}_3\right]^2}{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}$
$=\frac{[1.90]^2}{[0.60]^2[0.82]}=12.22 M^{-1}($ approx $)$

Question 6.3 At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms

I(g)⇌2I(g)

Calculate Kp for the equilibrium.

Answer :

It is given that total pressure (P T )is 10⁵ Pa and partial pressure of I atom = 40 % of P T
So, the partial pressure of I atom $=\frac{40}{100} \times 10^5=4 \times 10^4$

The partial pressure of I₂ = 60% of P T

So, the partial pressure of I2$=\frac{60}{100} \times 10^5=6 \times 10^4$

Now, for the reaction $I_{(g)} \rightleftharpoons 2 I_{(g)}$

$
\begin{aligned}
& K_p=\frac{\left(p_I\right)^2}{p_{l_2}} \\
& =\frac{\left(4 \times 10^4\right)^2}{6 \times 10^4} \mathrm{~Pa} \\
& =2.67 \times 10^4 \mathrm{~Pa}
\end{aligned}
$

Question 6.4 Write the expression for the equilibrium constant, K_c, for each of the following reactions:

(i) $2 \mathrm{NOCl}_{(g)} \rightleftharpoons 2 \mathrm{NO}_{(g)}+\mathrm{Cl}_{(g)}$
(ii) $2 \mathrm{Cu}\left(\mathrm{NO}_3\right)_{2(\mathrm{~s})} \rightleftharpoons 2 \mathrm{CuOS}_{(\mathrm{s})}+4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$
(iii) $\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(a q)}+\mathrm{H}_2 \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}_{(a q)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(a q)}$
(iv) $\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(a q)}+\mathrm{H}_2 \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOH}_{(a q)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(a q)}$
(v) $I_{2(s)}+5 F_2 \rightleftharpoons 2 I F_5$

Answer :

The equilibrium constant for any reaction can be written as (concentration of products) / (concentration of reactants). And we considered constant values for the solids and liquids because their density per unit volume or mass per unit volume does not change.
Thus,
(i)$K_c=\frac{[\mathrm{NO}]^2\left[\mathrm{Cl}_2\right]}{[\mathrm{NOCl}]^2}$

(ii)
$K_c=\frac{[\mathrm{CuO}]^2\left[\mathrm{NO}_2\right]^4\left[\mathrm{O}_2\right]}{\left[\mathrm{Cu}\left(\mathrm{NO}_3\right)_2\right]^2}$
(iii)
$K_c=\frac{\left[\mathrm{CH}_3 \mathrm{COOH}\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]}{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H} 5\right]}$

(iv)$K_C=\frac{1}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^{-}\right]}$

(v)$K_c=\frac{\left[I F_5\right]^2}{\left[F_2\right]^5}$

Question 6.5 Find out the value of K_c for each of the following equilibria from the value of K_p:

(i) $2 \mathrm{NOCl}_{(g)} \rightleftharpoons 2 \mathrm{NO}_{(g)}+\mathrm{Cl}_{2(g)} ; K_p=1.8 \times 10^{-2}$ at500K
(ii) $\mathrm{CaCO}_{3(s)} \rightleftharpoons \mathrm{CaO}_{(s)}+\mathrm{CO}_{2(g)} ; \mathrm{K}_p=167$ at 1073 K

Answer :

We know that the relation between Kp and Kc is expressed as;

$
K_p=K_c(R T)^{\Delta n}
$

here Δn = (no. of moles of product) - (no. of moles of reactants)

R = 0.0831 bar L /mol/K, and

For (i)
$K_p=1.8 \times 10^{-2}$and Temp (T) = 500K

Δn = 3 - 2 = 1
By putting the all values in eq (i) we get

$K_c=\frac{167}{0.0831 \times 1073}=1.87$

For (ii)
Kp = 167 and temp(T) = 1073 K
Δn = 2 - 1 = 1
Now, by putting all values in eq (i) we get,

$K_c=\frac{167}{0.0831 \times 1073}=1.87$

Question 6.6 For the following equilibrium, K_c = 6.3 × 10¹⁴ at 1000 K

$\mathrm{NO}_{(g)}+\mathrm{O}_{3(g)} \rightleftharpoons \mathrm{NO}_{2(g)}+\mathrm{O}_{2(g)}$

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?

Answer :

It is given that,
$K_c=6.3 \times 10^{14}$
we know that Kc′ for the reverse reaction is the inverse of the forward equilibrium constant. Thus it can be calculated as:

$\begin{aligned} & K_c^{\prime}=\frac{1}{K_c} \\ & =\frac{1}{6.3 \times 10^{14}} \\ & =1.58 \times 10^{-15}\end{aligned}$

Question 6.7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Answer :

For the pure liquids and solids, the molecular mass and the density at a particular temperature is always fixed and it is considered as a constant. Thus they can be ignored while writing the equilibrium constant expression

Question 6.8 Reaction between N₂ and O₂^- takes place as follows:

$2 \mathrm{~N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_{(g)}$

If a mixture of 0.482 mol N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which $K_c=2.0 \times 10^{-37}$ , determine the composition of equilibrium mixture.

Answer :

It is given that,
$K_c=2.0 \times 10^{-37}$
Let the concentration of N₂O at equilibrium be x . $$
\begin{array}{lcccc}
& 2 \mathrm{~N}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_{(g)} & \\
\text { initial conc } & 0.482 & 0.933 & 0 & \text { (in moles) } \\
\text { at equilibrium } & 0.482-x & 0.933-x & x & \text { (in moles) }
\end{array}
$$

The equilibrium constant is very small. So, we can assume $0.482-x=0.482$ and $0.933-x=0.933$

We know that,

$K_c=\frac{\left[\mathrm{N}_2 \mathrm{O}\right]^2}{\left[\mathrm{~N}_2\right]^2\left[\mathrm{O}_2\right]}$
$2 \times 10^{-37}=\frac{(x / 10)^2}{(0.0482)^2(0.0933)}$ {dividing the moles by 10 to get concentration of ions)

$\frac{x^2}{100}=2 \times 10^{-37} \times(0.0482)^2 \times(0.0933) x^2$

$
=43.35 \times 10^{-40} x
$

$
=\sqrt{43.35 \times 10^{-40}} x=6.6 \times 10^{-20}
$

So, the concentration of $\left[\mathrm{N}_2 \mathrm{O}\right]=\frac{x}{10}=6.6 \times 10^{-20}$

Question 6.9 Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:

$2 \mathrm{NO}_{(g)}+\mathrm{Br}_{2(g)} \rightleftharpoons 2 \mathrm{NOBr}_{(g)}$

When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate the equilibrium amount of NO and Br₂.

Answer :

The initial concentration of NO and Br₂ is 0.087 mol and 0.0437 mol respectively.

The given chemical reaction is-$2 \mathrm{NO}_{(g)}+\mathrm{Br}_{2(g)} \rightleftharpoons 2 \mathrm{NOBr}_{(g)}$

Here, 2 mol of NOBr produces from 2 mol of NO . So, 0.0518 mol of NOBr is obtained from 0.518 mol of NO .

Again, From 1 mol of Br₂ two mol of NOBr produced. So, to produce 0.518 mol of NOBr we need $\frac{0.518}{2}=0.0259 \mathrm{~mol}$ of $B r_2$.

Thus, the amount of NO present at equilibrium = 0.087 - 0.0518 = 0.0352 mol

and the amount of Br₂ present at the equilibrium = 0.0437-0.0259 = 0.0178 mol

Question 6.10 At 450K, Kp = 2.0 × 10¹⁰ /bar for the given reaction at equilibrium.

$2 \mathrm{SO}_{2(g)}+\mathrm{O}_{2(g)} \rightleftharpoons 2 \mathrm{SO}_{3(g)}$

What is K_c at this temperature?

Answer :

We have,
$K_p=2 \times 10^{10} /$ bar
We know that the relation between Kp and Kc ;

$K_p=K_c(R T)^{\Delta n}$
Here Δn = ( moles of product) - (moles of reactants)

$2 \mathrm{SO}_{2(g)}+O_{2(g)} \rightleftharpoons 2 \mathrm{SO}_{3(g)}$

So. here Δn = 2-3 = -1

By applying the formula we get;

$2 \times 10^{10}=K_c(0.0831 \mathrm{Lbar} / \mathrm{K} / \mathrm{mol}) \times 450 \mathrm{~K}$$K_c=\frac{2 \times 10^{10}}{(0.0831 \mathrm{Lbar} / \mathrm{K} / \mathrm{mol}) \times 450 \mathrm{~K}}$

$\begin{aligned} & =74.79 \times 10^{10} \mathrm{Lmol}^{-1} \\ & =7.48 \times 10^{10} \mathrm{Lmol}^{-1}\end{aligned}$

Question 6.11 A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?

$2 H I_{(g)} \rightleftharpoons H_{2(g)}+I_{2(g)}$

Answer :

The initial pressure of HI is 0.2 atm . At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 - 0.04 = 0.16 .

The given reaction is:

$\begin{gathered}2 H I_{(g)} \rightleftharpoons H_{2(g)}+I_{2(g)} \\ H I_{(g)} \rightleftharpoons 1 / 2 H_{2(g)}+1 / 2 I_{2(g)}\end{gathered}$

At equilibrium,

$\begin{aligned} & \text { pHI }=0.04 p H_2=\frac{0.16}{2} \\ & =0.08 p I_2=\frac{0.16}{2}=0.08\end{aligned}$

Therefore,

$\begin{aligned} & K_p=\frac{p H_2 \times p I_2}{p^2 H I} \\ & =\frac{0.08 \times 0.08}{(0.04)^2}=4\end{aligned}$

Question 6.12 A mixture of 1.57 mol of N₂ , 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction N_{2( g)}+3H_{2( g)}⇌2NH_{3( g)} is 1.7 × 10². Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer :

We have,$\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(g)}$

K_c=1.7×10²

The concentration of species are-$\left[\mathrm{N}_2\right]=1.57 / 20 \mathrm{molL}^{-1}\left[\mathrm{H}_2\right]=1.92 / 20 \mathrm{molL}^{-1}\left[\mathrm{NH}_3\right]=8.13 / 20 \mathrm{molL}^{-1}$

We know the formula of

$\begin{aligned} & Q_c=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3} \\ & =\frac{(8.13 / 20)^2}{(1.57 / 20)(1.92 / 20)^3}=2.4 \times 10^{-3}\end{aligned}$
The reaction is not in equilibrium. Since $Q c>K_c$, the equilibrium proceeds in reverse direction.

Question 6.14 One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,

$\mathrm{H}_2 \mathrm{O}_{(g)}+\mathrm{CO}_{(g)} \rightleftharpoons \mathrm{H}_{2(g)}+\mathrm{CO}_{2(g)}$

Calculate the equilibrium constant for the reaction.

Answer :

The given reaction is-

$\begin{array}{ccccc}\mathrm{H}_2 \mathrm{O}_{(g)}+\mathrm{CO}_{(g)} \rightleftharpoons \mathrm{H}_{2(g)}+\mathrm{CO}_{2(g)} & & & & \\ \text { initial conc } & 1 / 10 & 1 / 10 & 0 & 0 \\ \text { At equilibrium } & 0.6 / 10 & 0.6 / 10 & 0.04 & 0.04\end{array}$

Now, the equilibrium constant for the reaction can be calculated as;

$\begin{aligned} & K_C=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{CO}_2\right]}{\left[\mathrm{H}_2 \mathrm{O}\right][\mathrm{CO}]} \\ & =\frac{.04 \times .04}{(.06)^2}\end{aligned}$

= 0.44 (approx)

Question 6.15 At 700 K, equilibrium constant for the reaction:

$H_{2(g)}+I_{2(g)} \rightleftharpoons 2 H I_{(g)}$

is 54.8. If 0.5 mol L -1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H₂ (g) and I₂ (g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

Answer :

We have,

The equilibrium constant of the reaction = 54.8

moles of HI = 0.5 mol/L

The given reaction is-

$H_{2(g)}+I_{2(g)} \rightleftharpoons 2 H I_{(g)}$

So, the reverse equilibrium constant is$K_c^{\prime}=1 / K_c$

Suppose the concentration of hydrogen and iodine at equilibrium be x

$\left[I_2\right]=\left[H_2\right]=x$

Therefore,

$K_c^{\prime}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}{[\mathrm{HI}]}=\frac{x^2}{(.5)^2}$
So, the value of$x=\sqrt{\frac{0.25}{54.8}}=0.0675($ approx $) \mathrm{mol} / L$

Question 6.16 What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ?

$2 I C l_{(g)} \rightleftharpoons I_{2(g)}+C l_{2(g)} ; K_c=0.14$

Answer :

The given reaction is:$$
\begin{array}{lcl}
& 2 \operatorname{ICl}(g) \rightleftharpoons I_2(g)+C l_2(g) \\
\text { Initial conc. } & 0.78 \mathrm{M} \quad 0 \quad 0 \\
\text { At equilibrium } & (0.78-2 x) \mathrm{M} \quad x \mathrm{M} \quad x \mathrm{M}
\end{array}
$$

The value of Kc = 0.14

Now we can write,$\begin{aligned} & K_c=\frac{\left[I_2\right]\left[\mathrm{Cl}_2\right]}{[\mathrm{ICl}]^2} \\ & 0.14=\frac{x^2}{(.78-x)^2} 0.374\end{aligned}$

$=\frac{x}{(.78-x)}$

By solving this we can get the value of $x=0.167$

Question 6.17 Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

$C_2 H_{6(g)} \rightleftharpoons C_2 H_{4(g)}+H_{2(g)}$

Answer :

Suppose the pressure exerted by the hydrogen and ethene gas be p at equilibrium.

the given reaction is-

$\begin{array}{cllll}\mathrm{C}_2 \mathrm{H}_{6(\mathrm{~g})} \rightleftharpoons \mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} & & & & \\ \text { initial pressure } & 4 a t m & 0 & 0 \\ \text { At equilibrium } & 4-p & p & p\end{array}$

Now,

$\begin{aligned} & K_p=p_{C_2 H_4} \times p_{H_2} / p_{C_2 H_6} \\ & 0.04=\frac{p^2}{4-p}\end{aligned}$

By solving the quadratic equation we can get the value of p = 0.38

Hence,at equilibrium,

$p_{C_2 H_6}=4-\mathrm{p}=4-.038$

= 3.62 atm

Question 6.18(i) Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

$\mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)}$

(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)

Answer :

The given reaction is-
$\mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)}$ the concentration ratio (reaction quotient) of the given chemical reaction is-

$Q_c=\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5\right]\left[\mathrm{H}_2 \mathrm{O}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right]}$

Question 6.18(ii) Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

$\mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)}$

(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

Answer :

Let the volume of the mixture will be V.

$\begin{array}{cccc}\mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOO}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)} \\ \text { initial conc. } & 1 / \mathrm{V} & 0.18 / \mathrm{V} & 0 \\ \text { At equilibrium } & \frac{1-.171}{V}(=0.829 / \mathrm{V}) \frac{1-.171}{V} & 0.171 & 0.171\end{array}$

So equilibrium constant for the reaction can be calculated as;

$\begin{gathered}K_c=\frac{(0.171)^2}{(0.829)(0.009)} \\ =3.92 \text { (approx) }\end{gathered}$

Question 6.18(iii) Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

$$
\begin{array}{lccc}
& \mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)} \\
\text { initial conc. } & 1 / \mathrm{V} & 0.5 / \mathrm{V} & 0 \\
\text { At equilibrium } & \frac{1-.214}{V}(=0.786 / \mathrm{V}) \frac{1-.214}{V} & 0.214 & 0.214
\end{array}
$$

(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

Answer :

Let the volume of the mixture will be V.

$\begin{array}{cccc}\mathrm{CH}_3 \mathrm{COOH}_{(l)}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(l)} \rightleftharpoons \mathrm{CH}_3 \mathrm{COOO}_2 \mathrm{H}_{5(l)}+\mathrm{H}_2 \mathrm{O}_{(l)} & & & \\ \text { initial conc. } & 1 / \mathrm{V} & 0.18 / \mathrm{V} & 0 \\ \text { At equilibrium } & \frac{1-.171}{V}(=0.829 / \mathrm{V}) \frac{1-.171}{V} & 0.171 & 0.171\end{array}$

Therefore the reaction quotient of the reaction-

$\begin{aligned} & Q_c=\frac{(0.214)^2}{(0.786)(0.286)} \\ & =0.2037 \text { (approx) }\end{aligned}$

Since $Q_c<K_c$ equilibrium has not been reached.

Question 6.19 A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 × 10^-1 mol L ^-1 . If value of K_c is 8.3 × 10 ^-3 , what are the concentrations of PCl₃ and Cl₂ at equilibrium?

$P C l_{5(\mathrm{~g})} \rightleftharpoons P C l_{3(\mathrm{~g})}+C l_{2(\mathrm{~g})}$

Answer :

We have,

concentration of$P C l_5=0.05 \mathrm{~mol} / \mathrm{L}$
and $K_c=8.3 \times 10^{-3}$

Suppose the concentrations of both PCl₃ and Cl₂ at equilibrium be x mol/L. The given reaction is:

$
\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g)
$

$\begin{array}{llll}\text { at equilibrium } & \quad0.05 & \quad x & \quad x\end{array}$

it is given that the value of the equilibrium constant,$K_c=8.3 \times 10^{-3}$

Now we can write the expression for equilibrium as:

$\begin{aligned} & K_c=\frac{x^2}{0.05}=8.3 \times 10^{-3} \\ & \Rightarrow x=\sqrt{4.15 \times 10^{-4}} \\ & =0.0204 \text { (approx) }\end{aligned}$

Hence the concentration of$\mathrm{PCl}_3$ and $\mathrm{Cl}_2$ is $0.0204 \mathrm{~mol} / \mathrm{L}$

Question 6.20 One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO₂ . $F e O_{(s)}+C O_{(g)} \rightleftharpoons F e_{(s)}+C O_{2(g)}$; Kp = 0.265 atm at 1050K What are the equilibrium partial pressures of CO and $\mathrm{CO}_2$ at 1050 K if the initial partial pressures are: pCO = 1.4 atm and $p \mathrm{CO}_2$ = 0.80 atm?

Answer :

We have,

$K_p=0.265$
the initial pressure of CO and CO₂ are 1.4 atm and 0.80atm resp.

The given reaction is-

$\mathrm{FeO}_{(s)}+\mathrm{CO}_{(g)} \rightleftharpoons \mathrm{Fe}_{(s)}+\mathrm{CO}_{2(g)}$

initially, 1.4 atm 0.80 atm

$Q_p=\frac{p C O_2}{p C O}=\frac{0.80}{1.4}=0.571$
Since Qp>Kc the reaction will proceed in the backward direction to attain equilibrium. The partial pressure of CO₂ will increase = decrease in the partial pressure of CO₂ = p

$\therefore K_p=\frac{p C O_2}{p C O}$

$\begin{aligned} & =\frac{0.80-p}{1.4+p}=0.265 \\ & =0.371+.265 p=0.80-p\end{aligned}$

By solving the above equation we get the value of p = 0.339 atm

Question 6.21 Equilibrium constant, K_c for the reaction

$\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(g)}$

At 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L^-1 N_{2 ,} 2.0 mol L^-1 H₂ and 0.5 mol L^-1 NH₃. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Answer :

The given reaction is:

$$\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightleftharpoons 2 \mathrm{NH}_{3(g)} \\$$ $$\quad 3.0 \mathrm{molL}^{-1} 2.0 \mathrm{molL}^{-1} 0.5 \mathrm{molL}^{-1}$$

Now, we know that,

$\begin{aligned} & Q c=\left[\mathrm{NH}_3\right] 2^{/}\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3 \\ & =(0.5)^2 /(3.0)(2.0)^3=0.0104\end{aligned}$

It is given that Kc = 0.061

Since, Qc ≠ Kc, the reaction mixture is not at equilibrium.

Again, Qc<Kc , the reaction will proceed in the forward direction to attain the equilibrium.

Question 6.22 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

$2 \mathrm{BrCl}_{(g)} \rightleftharpoons \mathrm{Br}_{2(g)}+\mathrm{Cl}_{2(g}$

For which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10^-3 mol L^-1 , what is its molar concentration in the mixture at equilibrium?

Answer :

Suppose the x amount of bromine and chlorine formed at equilibrium. The given reaction is:

$$
\begin{aligned}
& \qquad 2 \mathrm{BrCl}_{(g)} \rightleftharpoons \mathrm{Br}_{2(g)}+\mathrm{Cl}_{2(g)} \\
& \text { Initial Conc. } 3.3 \times 10^{-3} \quad 0 \quad 0 \\
& \text { at equilibrium } 3.3 \times 10^{-3.2 x} \quad x \quad x
\end{aligned}
$$

Now, we can write,

$\begin{aligned} & K_c=\frac{\left[\mathrm{BR}_2\right]\left[\mathrm{Cl}_2\right]}{\left[\mathrm{BrCl}_2\right]^2} \\ & 32=\frac{x^2}{\left(3.3 \times 10^{-3}-2 x\right)^2} \\ & 5.66=\frac{x}{\left(3.3 \times 10^{-3}-2 x\right)} \\ & x+11.32 x=18.678 \times 10^{-3}\end{aligned}$

By solving the above equation we get,
$x=1.51 \times 10^{-3}$

Hence, at equilibrium$\left[\mathrm{BrCl}_2\right]=3.3 \times 10^{-3}-\left(2 \times 1.51 \times 10^{-3}\right)$
=0.3×10−3 M

Question 6.24 Calculate a) ΔG⁰ for the formation of NO₂ from NO and O₂ at 298K

$\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g)$

where

$\begin{aligned} & \Delta_f G^{+}\left[\mathrm{NO}_2\right]=52.0 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta_f G^{+}[\mathrm{NO}]=87.0 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta_f G^{+}\left[\mathrm{O}_2\right]=0 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$

Answer :

Given data,

$\begin{aligned} & \Delta_f G^{+}\left[\mathrm{NO}_2\right]=52.0 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta_f G^{+}[\mathrm{NO}]=87.0 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta_f G^{+}\left[\mathrm{O}_2\right]=0 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$

Given chemical reaction

-$\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g)$

for the reaction,$\begin{aligned} & \Delta G^0=\Delta G^0 \text { (products) }-\Delta G^0 \text { (reactants) } \\ & =(52-87-0) \\ & =-35 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$

Question 6.25 Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

(a) $\mathrm{PCl}_{5(g)} \rightleftharpoons \mathrm{PCl}_{3(g)}+\mathrm{Cl}_{2(g)}$
(b) $\mathrm{CaO}(s)+\mathrm{CO}_2(s) \rightleftharpoons \mathrm{CaCO}_3(s)$
(c) $3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_3 \mathrm{O}_4(\mathrm{~s})+\mathrm{H}_2(\mathrm{~g})$

Answer :

According to Le Chatellier's principle, if the pressure is decreased, then the equilibrium will shift in the direction in which more number of moles of gases is present.

So,

• (a)The number of moles of reaction products will increase.
• (b)The number of moles of reaction products will decrease
• (c)The number of moles of reaction products remains the same.

Question 6.26 Which of the following reactions will get affected by increasing the pressure? Also, mention whether the change will cause the reaction to go into forward or backward direction.

(i) $\mathrm{COCl}_2(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_2(g)$
(ii) $\mathrm{CH}_4(g)+2 \mathrm{~S}_2(g) \rightleftharpoons \mathrm{CS}_2(g)+2 \mathrm{H}_2 \mathrm{~S}(g)$
(iii) $\mathrm{CO}_2(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})$
(iv) $2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$

(v) $\mathrm{CaCO}_3(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2(g)}$

(vi)$4 \mathrm{NH}_3(g)+5 \mathrm{O}_2(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_2 \mathrm{O}(g)$

Answer :

According to Le Chatellier's principle, if the pressure is increased, then the equilibrium will shift in the direction in which less number of moles of gases is present. So, as per this rule following given reactions are affected by the increasing pressure-

The reaction (i), (iii), and (vi)- all proceeds in the backward direction

Reaction(iv) will shift in the forward direction because the number of moles of gaseous reactants is more than that of products.

Question 6.27 The equilibrium constant for the following reaction is 1.6 ×10⁵ at 1024K

H_2(g)+Br_2(g)⇌2HBr_(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.

Answer :

Given that,

$\begin{aligned} & K_p \text { for the reaction }=1.6 \times 10^5 \\ & K_p^{\prime}=1 / K_p=\frac{1}{1.6 \times 10^5}=6.25 \times 10^{-6}\end{aligned}$

Let the pressure of both H₂ and Br₂ at equilibrium be p.

$\begin{array}{cccc}2 \mathrm{HBr} & \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g}) & & \\ \text { initial conc. } & 10 & 0 & 0 \\ \text { at eq } & 10-2 \mathrm{p} & \mathrm{p} & \mathrm{p}\end{array}$

Now,
$K_p^{\prime}=\frac{p^{B r_2} \cdot p^{H_2}}{p_{H B r}^2}$
$\begin{aligned} & 6.25 \times 10^{-6}=\frac{p^2}{(10-2 p)^2} \\ & 5 \times 10^{-3}=\frac{p}{(5-p)}\end{aligned}$
By solving the above equation we get,

p = 0.00248 bar

Hence the pressure of H₂ and Br₂ is 0.00248 bar and pressure of HBr is 0.00496 bar

Question 6.28(b) Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

$\mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g})$

How will the values of Kp and composition of equilibrium mixture be affected by

(i) increasing the pressure

(ii) increasing the temperature

(iii) using a catalyst?

Answer :

(i) According to Le Chatellier's principle, if pressure is increased, then the reaction will shift towards the less number of moles of gases. So, here the direction of equilibrium is backward and the value of Kp decreases.

(ii) According to Le Chatellier's principle, as the reaction is endothermic, the equilibrium will shift in the forward direction. The value of Kp is increases.

(iii) The equilibrium of the reaction is not affected by the presence of the catalyst. It only increases the rate of reaction.

Question 6.29(a) Describe the effect of :

(a) addition of H₂

$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$

Answer :

$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$
(a)According to Le Chatelliers principle, on the addition of dihydrogen, the number of mole of H₂ increases on the reactant side. Thus to attain the equilibrium again the reaction will move in the forward direction.

(b) According to Le Chatellier's principle, on the addition of methyl alcohol, the number of moles of methyl alcohol increases on the product sides. So, to attain the equilibrium, the reaction will proceed in a backward direction.

(c) If we remove the CO from the reactant side, the concentration on the reactant side will decrease and to attain an equilibrium, the reaction will shift backward direction

(d) On removal of CH₃OH the equilibrium will shift in the forward direction.

Question 6.29(b) Describe the effect of :

addition of CH₃OH

$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$

Answer :

$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$
According to Le Chatellier's principle, on the addition of methyl alcohol, the number of moles of methyl alcohol increases on the product sides. So, to attain the equilibrium, the reaction will proceed in a backward direction.

Question 6.29(c) Describe the effect of :

removal of CO

$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$

Answer :

$2 \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})$
If we remove the CO from the reactant side, the concentration on the reactant side will decrease and to attain an equilibrium, the reaction will shift backward direction

Question 6.30 At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl₅ is 8.3 ×10 ^-3 . If decomposition is depicted as,

$\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g), \Delta_r \mathrm{H}^{+}=124.0 \mathrm{kJmol}^{-1}$

(a) write an expression for Kc for the reaction.
(b) what is the value of Kc for the reverse reaction at the same temperature ?
(c) what would be the effect on Kc if

(i) more PCl₅ is added
(ii) pressure is increased
(iii) the temperature is increased ?

Answer :

We have,
$\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g), \Delta_r \mathrm{H}^{+}=124.0 \mathrm{kJmol}^{-1}$
Equilibrium constant for the above reaction = 8.3×10⁻³

(a) Expression of Kc for this reaction-$K_c=\frac{\left[P C l_3\right]\left[C l_2\right]}{P C l_5}$

(b) The value of the reverse equilibrium constant can be calculated as;

$\begin{aligned} & K_c^{\prime}=\frac{1}{K_c} \\ & =\frac{1}{8.3 \times 10^{-3}}=1.20 \times 10^2\end{aligned}$

(c). i Kc would remain the same because the temperature is constant in this case.

(c). ii If we increase the pressure, there is no change in Kc because the temperature is constant in this case also.

(c). iii In an endothermic reaction, the value of Kc increases with an increase in temperature.

Question 6.31 Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂ . In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

$\mathrm{CO}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_2+\mathrm{H}_2(\mathrm{~g})$

If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and steam such that $p_{\mathrm{CO}}=p_{\mathrm{H}_2 \mathrm{O}}=4.0 \mathrm{bar}$, what will be the partial pressure of H₂ at equilibrium? Kp= 10.1 at 400°C

Answer :

We have,
The partial pressure of CO and H₂O is 4 bar and the Kp=10.1
Let p be the partial pressure of CO₂ and H₂ at equilibrium. The given reaction is-
$$
\begin{array}{lcccc}
& \mathrm{CO}(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{~g}) & \rightleftharpoons \mathrm{CO}_2+\mathrm{H}_2(\mathrm{~g}) \\
\text { Initial concentration } & 4 \text { bar } & 4 \text { bar } & 0 & 0 \\
\text { At equilibrium } & 4-p & 4-p & p & p
\end{array}
$$

Therefore, we can write,$
\begin{aligned}
& 10.1=\frac{p_{\mathrm{CO}_2} \cdot p_{\mathrm{H}_2}}{p_{\mathrm{CO}} \cdot p_{\mathrm{H}_2 \mathrm{O}}} \\
& 10.1=\frac{p^2}{(4-p)^2} 3.178 \\
& =\frac{p}{(4-p)} p=(4-p) \times 3.178
\end{aligned}
$
By solving the above equation we get, p = 3.04

Hence, the partial pressure of dihydrogen at equilibrium is 3.04 bar

Question 6.32 Predict which of the following reaction will have an appreciable concentration of reactants and products:

(a) $\mathrm{Cl}_2(g) \rightleftharpoons 2 \mathrm{Cl}(g) ; K_c=5 \times 10^{-39}$
(b) $\mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g}) ; K_c=3.7 \times 10^8$
(c) $\mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{NO}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2 \mathrm{Cl}(\mathrm{g}) ; \mathrm{K}_c=1.8$

Answer :

If the value of Kc is in the range of 10⁻³ to 10³, then the reaction has appreciable concentrations of both reactants and products.

Therefore, the third reaction (c) (K_c=1.8) will have an appreciable concentration of reactants and products.

Question 6.33 The value of K c for the reaction 3O_2(g)⇌2O_3(g) is 2.0 ×10^-50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 ×10^-2 , what is the concentration of O₂ ?

Answer :

We have,

Equilibrium constant of the reaction = 2×10⁻⁵⁰
the concentration of dioxygen$\left[O_2\right]=1.6 \times 10^{-2}$

The given reaction is-

$3 O_2(g) \rightleftharpoons 2 O_3(g)$
Then we have,
(equilibrium constant)
$K_c=\left[O_3(g)\right]^2 /\left[O_2(g)\right]^3$
$\begin{aligned} & 2 \times 10^{-50}=\left[\mathrm{O}_3\right]^2 /\left(1.6 \times 10^{-2}\right)^3\left[\mathrm{O}_3\right]^2 \\ & =2 \times 10^{-50} \times\left(1.6 \times 10^{-2}\right)^3 \\ & =8.192 \times 10^{-56} \\ & {\left[\mathrm{O}_3\right]=\sqrt{8.192 \times 10^{-56}}=2.86 \times 10^{-28}}\end{aligned}$

Thus the concentration of dioxygen is$2.86 \times 10^{-28}$

Question 6.34(a) The reaction,

$\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})$

is at equilibrium at 1300 K in a 1L flask. It also contains 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O, and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant Kc, for the reaction at the given temperature is 3.90.

Answer :

Given that,
Total volume = 1L
0.3 mol of CO , 0.10 mol of dihydrogen( H₂ )and the 0.02 mol of water( H₂O )
the equilibrium constant = 3.90

Let x be the concentration of methane at equilibrium. The given reaction is-

$
\mathrm{CO}(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{~g})
$

At equilibrium, $\quad 0.3 \mathrm{~mol} / \mathrm{L} \quad 0.1 \mathrm{~mol} / \mathrm{L} \quad x \quad 0.02 \mathrm{~mol} / \mathrm{L}$

Therefore,

$\begin{aligned} & K_c=\frac{0.02 \times x}{(0.3)(0.1)}=3.90 \\ & \Rightarrow x=\frac{(3.9)(0.3)(0.1)^3}{(0.02)} \\ & =5.85 \times 10^{-2}\end{aligned}$

Thus the concentration of methane at equilibrium is$=5.85 \times 10^{-2}$

Question 6.35 What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:

$\mathrm{HNO}_2, \mathrm{CN}^{-}, \mathrm{HClO}_4, \mathrm{~F}^{-}, \mathrm{OH}^{-}, \mathrm{CO}_3^{-2}$ and $S^{2-}$

Answer :

A conjugate acid-base pair means that the species are differed by only one proton. for example; HCl is an acid because it donates a proton to water. So $\mathrm{HCl}-\mathrm{Cl}^{-}$, and $\mathrm{H}_3 \mathrm{O}^{+}-\mathrm{H}_2 \mathrm{O}$ these pairs are called conjugate acid-base pair.

Question 6.36 Which of the followings are Lewis acids?$\mathrm{H}_2 \mathrm{O}, \mathrm{BF}_3, \mathrm{H}^{+}$and $\mathrm{NH}_4^{+}$

Answer :

Lewis acid-
Those species that can accept the pair of electrons are called Lewis acids. For example Boron trifluoride ( BF₃ ), ammonium ion( NH₄⁺) and the hydrogen ion( H⁺). Among them water molecule is a Lewis base, it can donate pair of electrons.

Question 6.37 What will be the conjugate bases for the Brönsted acids: $\mathrm{HF}, \mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{HCO}^{-3}$?

Answer :

When Brönsted acids lose their proton then they become a conjugate base of that corresponding acids.
Followings are the conjugate base of Brönsted acid-

$\begin{aligned} & \mathrm{HF}_2-F^{-} \\ & \mathrm{H}_2 \mathrm{SO}_4^{-}-\mathrm{HSO}_4^{-} \\ & \mathrm{HCO}_3^{-}-\mathrm{CO}_3^{2-}\end{aligned}$

Question 6.38 Write the conjugate acids for the following Brönsted bases: NH₂⁻,NH₃ and HCOO⁻ .

Answer :

When Brönsted base accepts a proton then they become a conjugate acid of that corresponding base.
Followings are the conjugate acid of Brönsted base-

• $\mathrm{NH}_2{ }^{-}-\mathrm{NH}_3$
• $\mathrm{NH}_3-\mathrm{NH}_4{ }^{+}$
• $\mathrm{HCOO}^{-}-\mathrm{HCOOH}$

Question 6.39 The species: H2O,HCO3−,HSO4−andNH3 can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

Answer :

When acid or base accept or lose a proton, they form conjugate acid or base of that corresponding species.

Lists of the conjugate acid and conjugate base of the given species-

Question 6.40 Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:
(a) OH⁻
(b) F⁻
(c) H⁺ .
(d) BCl₃

Answer :

Species which donate pair of an electron are called Lewis base and which accepts pair of electrons are called acid.

(a) OH⁻ is a Lewis base since it can donate its lone pair of electrons.

(b) F⁻ is a Lewis base since it can donate a pair of electrons.

(c) H⁺ is a Lewis acid since it can accept a pair of electrons.

(d) BCl₃ is a Lewis acid since it can accept a pair of electrons.

Question 6.41 The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3 M. what is its pH?

Answer :

We have,
the concentration of Hydrogen ion sample is 3.8×10−3 M

So, $p H=-\log [H]^{+}$

$
=-\log \left(3.8 \times 10^{-3}\right)=3-\log 3.8=2.42
$

Question 6.42 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

Answer :

We have,
The pH of a sample of vinegar is 3.76$p H=?$

Therefore,
$p H=-\log \left[H^{+}\right] \log \left[H^{+}\right]=-p H$
Taking antilog on both sides we get,

$\begin{aligned} & {\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-3.76)} \\ & {\left[\mathrm{H}^{+}\right]=1.74 \times 10^{-4}}\end{aligned}$

Hence the concentration of hydrogen ion$\left[H^{+}\right]=1.74 \times 10^{-4}$

Question 6.43 The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10^-4 , 1.8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Answer :

We have,
IOnization constant of hydrogen fluoride, methanoic acid and hydrogen cyanide are $6.8 \times 10^{-4}, 1.8 \times 10^{-4}$ and $4.8 \times 10^{-9}$respectively.

It is known that,
$
K_b=\frac{K_w}{K_a}
$..........................(i)

$K_b$ of the conjugate base $F^{-}$

$
\begin{aligned}
& =\frac{10^{-14}}{6.8 \times 10^{-4}} \\
& =1.5 \times 10^{-11}
\end{aligned}
$

Similarly,
By using the equation (i)

$K_b$ of the conjugate base $\mathrm{HCOO}^{-}$

$
\begin{aligned}
& =\frac{10^{-14}}{1.8 \times 10^{-4}} \\
& =5.6 \times 10^{-11}
\end{aligned}
$

$K_b$ of the conjugate base $\mathrm{CN}^{-}$

$
\begin{aligned}
& =\frac{10^{-14}}{4.8 \times 10^{-9}} \\
& =2.8 \times 10^{-6}
\end{aligned}
$

Question 6.44 The ionization constant of phenol is 1.0 × 10 ^-10 . What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

Answer :

We have,
The ionization constant of phenol is $1.0 \times 10^{-10}$ , and
the concentration of phenol is 0.05 M
degree of ionisation = ?

Ionization of phenol;
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}+\mathrm{H}_3 \mathrm{O}^{+}$
At equilibrium,
the concentration of various species are-
$\begin{aligned} & {\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right]=0.05-x} \\ & {\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}\right]=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=x}\end{aligned}$
As we see, the value of ionisation is very less. Also x will be very small. Thus we can ignore x .

$\begin{aligned} & K_a=\frac{\left[C_6 \mathrm{H}_5 \mathrm{O}^{-}\right]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]}{\left[C_6 \mathrm{H}_5 \mathrm{OH}\right]} 10^{-10} \\ & =\frac{x^2}{0.05} x=\sqrt{10^{-10} \times 0.05} \\ & x=2.2 \times 10^{-6}\end{aligned}$

Hence the concentration of phenolate ion is$\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}\right]=2.2 \times 10^{-6}$

Let α be the degree of dissociation of phenol in the presence of 0.01 M of phenolate ion.

$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons \mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}+\mathrm{H}^{+}$
Concentration (1 - α ) 0.05 0.05 α 0.05 α

So,$\begin{aligned} & {\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right]=0.05(1-\alpha)=0.05} \\ & {\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}\right]=0.05 \alpha+0.01 \approx 0.01 \mathrm{M}} \\ & {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=0.05 \alpha}\end{aligned}$

therefore,

$K_a=\frac{(0.01)(0.05 \alpha)}{0.05} 10^{-10}=0.01 \alpha$

$\alpha=10^{-8}$
The degree of dissociation is 10−8

Question 6.45 The first ionization constant of H₂S is 9.1 × 10 ^-8 . Calculate the concentration of HS^- ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H₂S is 1.2 × 10 ^-13 , calculate the concentration of S^2- under both conditions.

Answer :

We have,
1st ionisation constant of hydrogen sulphide is 9.1×10⁻⁸ and the 2nd dissociation constant is 1.2×10⁻¹³

Case 1st-(absence of hydrochloric acid)

To calculate the concentration of HS⁻
Let x be the concentration of HS⁻ and the ionisation of hydrogen sulphide is;
$\mathrm{H}_2 \mathrm{~S} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HS}^{-}$
0.1 M

At equilibrium, the concentration of various species is,
Since the dissociation constant is very small. So, x can be neglected.
the concentration of $H_2 S=1-x \mathrm{M}$
the concentration of$H S^{-}$and $H^{+}$is $x \mathrm{M}$
So,
$K_a=\frac{x^2}{0.1}=9.1 \times 10^{-8}$
from here x can be calculated and we get, $x=9.54 \times 10^{-5} M$

Case 2nd (In presence of 0.1 M, HCl)
Suppose H₂S is dissociated is x .Then at equilibrium,

$\left[\mathrm{H}_2 \mathrm{~S}\right]=0.1-x \simeq 0.1,\left[\mathrm{H}^{+}\right]=0.1+x \simeq 0.1$ and the $\left[\mathrm{HS}^{-}\right]=x \mathrm{M}$
So,

$
K_a=\frac{x(0.1)}{0.1}=9.1 \times 10^{-8}
$
Thus the concentration of $\left[\mathrm{HS}^{-}\right]$is $9.1 \times 10^{-8}$

Question 6.46 The ionization constant of acetic acid is 1.74 × 10 ^-5 . Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Answer :

It is given,
The ionisation constant of acetic acid is 1.74×10⁻⁵ and concentration is 0.05 M

The ionisation of acetic acid is;

$\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$
Therefore,

$\begin{aligned} & \mathrm{K}_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}=\frac{\left[\mathrm{H}^{+}\right]^2}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ & \therefore\left[\mathrm{H}^{+}\right]=\sqrt{\left(1.74 \times 10^{-5}\right)\left(5 \times 10^{-2}\right)} \\ & =9.33 \times 10^{-4}=\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\end{aligned}$

$=9.33 \times 10^{-4}=\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]$

$
\begin{aligned}
& =4-\log (9.33) \\
& =3.03
\end{aligned}
$
We know that,

$
\begin{aligned}
\alpha & =\sqrt{\frac{K_a}{C}} \\
\alpha & =\sqrt{\frac{1.74 \times 10^{-5}}{0.05}} \\
& =1.86 \times 10^{-2}
\end{aligned}
$

Question 6.47 It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa .

Answer :

We have,
pH of organic acid is 4.15 and its concentration is 0.01M

Suppose the organic acid be HA. The dissociation of organic acid can be written as;

$\begin{aligned} & H A \rightleftharpoons H^{+}+A^{-} \\ & p H=4.15\end{aligned}$$-\log \left[H^{+}\right]=4.15\left[H^{+}\right]=7.08 \times 10^{-5}$

Now,

$\begin{aligned} & K_a=\left[H^{+}\right]\left[A^{-}\right] /[H A] \\ & {\left[H^{+}\right]=\left[A^{-}\right]=7.08 \times 10^{-5}}\end{aligned}$

[HA] = 0.01

Then,

$K_a=\left(7.08 \times 10^{-5}\right)^2 / 0.01 K_a=5.01 \times 10^{-7}$

Thus

$p K_a=-\log \left(5.01 \times 10^{-7}\right)=6.30$

Question 6.48 Assuming complete dissociation, calculate the pH of the following solutions:

(a) 0.003 M HCl

(b) 0.005 M NaOH

(c) 0.002 M HBr

(d) 0.002 M KOH

Answer :

Assuming the complete dissociation. So, α=1

(a) The ionisation of hydrochloric acid is
$\mathrm{HCl} \rightleftharpoons \mathrm{H}^{+}+\mathrm{Cl}^{-}$
Since it is fully ionised then$\left[\mathrm{H}^{+}\right]=\left[\mathrm{Cl}^{-}\right]=0.003 \mathrm{M}$
Therefore,$\begin{aligned} & p H \text { of the solution }=-\log (0.003) \\ & =3-\log (3) \\ & =2.52\end{aligned}$

(b) The ionisation of 0.005M NaOH
$\begin{aligned} & \mathrm{NaOH} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{OH}^{-} \\ & {\left[\mathrm{Na}^{+}\right]=\left[\mathrm{OH}^{-}\right]=0.005 \mathrm{M}}\end{aligned}$
Therefore,

pOH of the solution $
=-\log (0.005)=3-\log 5=2.301
$

pH of the solution is equal to (14 - 2.301 =11.70)

(c) The ionisation of 0.002MHBr
$\begin{aligned} & \mathrm{HBr} \rightleftharpoons \mathrm{H}^{+}+\mathrm{Br}^{-} \\ & {\left[\mathrm{Br}^{-}\right]=\left[\mathrm{H}^{+}\right]=0.002 \mathrm{M}}\end{aligned}$
Therefore,

$p H$ of the solution $
=-\log (0.002)=3-\log 2=2.69
$

pH of the solution is equal to (2.69)

(d) The ionisation of 0.002M KOH
$\begin{aligned} & \mathrm{KOH} \rightleftharpoons \mathrm{K}^{+}+\mathrm{OH}^{-} \\ & {\left[\mathrm{OH}^{-}\right]=\left[\mathrm{K}^{+}\right]=0.002 \mathrm{M}}\end{aligned}$
Therefore,

pOH of the solution $
=-\log (0.002)=3-\log 2=2.69
$

pH of the solution is equal to (14 - 2.69 = 11.31)

Question 6.49(a) Calculate the pH of the following solutions:

2 g of TlOH dissolved in water to give 2 litre of solution.

Answer :

Here, 2 g of TlOH dissolves in water to give 2 litres of solution
So, the concentration of TlOH =$[$ TlOH $($ aq $)]=\frac{2}{2} g / L=\frac{1}{221} M$ (the molar mass of TlOH is 221)

TlOH can be dissociated as $\mathrm{TlOH}(\mathrm{aq}) \rightarrow \mathrm{Tl}^{+}+\mathrm{OH}^{-}$

$
O H^{-}(a q)=\operatorname{TlOH}(a q)=\frac{1}{221} M
$

Therefore, $K_w=\left[H^{+}\right]\left[O H^{-}\right]\left(\right.$since $\left.\mathrm{K} w=10^{-14}\right)$
So, the concentration of $\left[H^{+}\right]=221 \times 10^{-14}$
Thus $P^H=-\log \left[H^{+}\right]=-\log \left(221 \times 10^{-14}\right)$
= 11.65(approx)

Question 6.49(b) Calculate the pH of the following solutions:

0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution.

Answer :

The calcium hydroxide ion dissociates into-

$
\mathrm{Ca}(\mathrm{OH})_2 \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-}
$

Molecular weight of $\mathrm{Ca}(\mathrm{OH})_2=74$
the concentration of $\left[\mathrm{Ca}(\mathrm{OH})_2\right]=\frac{0.3 \times 1000}{74 \times 500}=0.0081 \mathrm{M}$

$\therefore\left[\mathrm{OH}^{-}\right]=\left[\mathrm{Ca}(\mathrm{OH})_2\right]=0.0081 \mathrm{M}$

We know that,

$\begin{aligned} & {\left[H^{+}\right]=\frac{K_w}{\left[O H^{-}\right]}=\frac{10^{-14}}{0.0162}} \\ & =61.7 \times 10^{-14}\end{aligned}$

Thus $pH=-\log \left[H^{+}\right]=-\log \left(61.7 \times 10^{-14}\right)$

$=14-1.79=12.21$

Question 6.49(c) Calculate the pH of the following solutions:

0.3 g of NaOH dissolved in water to give 200 mL of solution.

Answer :

NaOH dissociates into$\mathrm{NaOH} \rightarrow \mathrm{Na}^{+}+\mathrm{OH}^{-}$
So, the concentration of $[\mathrm{NaOH}]=\frac{0.3 \times 1000}{200 \times 40} \mathrm{M}=0.0375 \mathrm{M}$

$\therefore\left[\mathrm{OH}^{-}\right]=[\mathrm{NaOH}]=0.0375 \mathrm{M}$

We know that ,

$
\left[H^{+}\right]=\frac{K_W}{\left[O H^{-}\right]}=\frac{10^{-14}}{0.0375}=2.66 \times 10^{-13}
$

Now, $pH=-\log \left[H^{+}\right]$

$
=-\log \left(2.66 \times 10^{-13}\right)=12.57
$

Question 6.49(d) Calculate the pH of the following solutions:

1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

Answer :

We know that,
M₁V₁ (before dilution) = M₂V₂ (after dilution)

initially V₁ = 1mL and M₁ = 13.6 M
and V₂ = 1L and M₁ = ?

By putting all these values we get,

$
\begin{aligned}
& M_2=\frac{13.6 \times 10^{-3}}{1}=1.36 \times 10^{-2} \\
& \therefore\left[H^{+}\right]=1.36 \times 10^{-2}
\end{aligned}
$
Thus $pH=-\log \left[H^{+}\right]=-\log \left(1.36 \times 10^{-2}\right)$

= 1.86 (approx)

Question 6.50 The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.

Answer :

We have,

Degree of ionization(a) = 0.132
Concentration of bromoacetic acid (C) = 0.1 M

Thus the concentration of H₃O⁺=C.a
= 0.0132

Therefore

$\begin{aligned} & pH=-\log \left[H^{+}\right]=-\log (0.0132) \\ & =1.879\end{aligned}$

Now, we know that,

$\begin{aligned} & K_a=C . a^2 \\ & \text { So, } p K_a=-\log \left(C . a^2\right) \\ & =-\log \left(0.1 \times(0.0132)^2\right) \\ & =-\log (0.0017)=2.76 \text { (approx) }\end{aligned}$

Question 6.51 The pH of 0.005M codeine ($C_{18} \mathrm{H}_{21} \mathrm{NO}_3$) solution is 9.95. Calculate its ionization constant and pK b.

Answer :

We have,

$\begin{aligned} & \mathrm{C}=0.005 \mathrm{M} \\ & pH=9.95 \text { and } pOH=14-9.95=4.05\end{aligned}$

we know that pOH=-$\log \left[O H^{-}\right]$$4.05=-\log \left[\mathrm{OH}^{-}\right]$

By taking antilog on both sides we get.

concentration of$\left[\mathrm{OH}^{-}\right]=8.91 \times 10^{-5}$

C.a $=8.91 \times 10^{-5}$

So, $\mathrm{a}=1.782 \times 10^{-2}$

We know that,

$
\begin{aligned}
& K_b=C . a^2 \\
& =0.005 \times\left(1.782 \times 10^{-2}\right)^2=0.0158 \times 10^{-4}
\end{aligned}
$

Thus

$
\begin{aligned}
& p K_b=-\log \left(K_b\right) \\
& =-\log \left(0.0158 \times 10^{-4}\right)=5.80
\end{aligned}
$

Question 6.52 What is the pH of 0.001M aniline solution? Calculate the degree of ionization of aniline in the solution. Also, calculate the ionization constant of the conjugate acid of aniline.(K_b = 4.27×10⁻¹⁰ )

Answer :

We have,
C = 0.001 M
K_b = 4.27×10⁻¹⁰
Degree of inozation of aniline (a) = ?
Ionization constant of the conjugate acid ( K_a ) = ?

We know that

$\begin{aligned} & K_b=C . a^2 \\ & 4.27 \times 10^{-10}=(0.001) a^2 \\ & \text { Thus } a=\sqrt{4270 \times 10^{-10}} \\ & =65.34 \times 10^{-4}\end{aligned}$

$\begin{aligned} & \text { Then [Base] }=\text { C.a }=\left(65.34 \times 10^{-4}\right)(0.001) \\ & =0.653 \times 10^{-5}\end{aligned}$

$\begin{aligned} & \text { Now, } pOH=-\log \left(0.65 \times 10^{-5}\right)=6.187 \\ & P^H=14-PO H \\ & =14-6.187 \\ & =7.813\end{aligned}$

It is known that,

$
K_a \times K_b=K_w
$

So, $K_a=\frac{10^{-14}}{4.27 \times 10^{-10}}$ $=2.34 \times 10^{-5}$ This is ionization constant.

Question 6.53(a) Calculate the degree of ionization of 0.05M acetic acid if its pK a value is 4.74. How is the degree of dissociation affected when its solution also contains

0.01M

Answer :

We have,
C = 0.05 M
$p K_a=4.74=-\log \left(K_a\right)$
By taking antilog on both sides we get,

$
\left(K_a\right)=1.82 \times 10^{-5}=C \cdot(\alpha)^2
$

from here we get the value of $\alpha=\sqrt{\frac{1.82 \times 10^{-5}}{5 \times 10^{-2}}}$ $=1.908 \times 10^{-2}$

After adding hydrochloric acid, the concentration of H+ ions increases and due to that the equilibrium shifts towards the backward direction. It means dissociation will decrease.

(i) when 0.01 HCl is taken

$\begin{array}{cccc}\mathrm{CH}_3 \mathrm{COOH} & \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} & & \\ \text {Initial conc. } & 0.05 & 0 & 0 \\ \text { after dissociation } & 0.05-x & 0.001+x & x\end{array}$

As the dissociation is very small.
So we can write $0.001+x \approx 0.001$ and $0.05-x \approx 0.05$

$\begin{aligned} & \text { Now, } K_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ & =\frac{(0.001)(x)}{(0.05)}=\frac{x}{50}\end{aligned}$

So, the value of $x=\frac{\left(1.82 \times 10^{-5}\right)(0.05)}{0.01}$
Now degree of dissociation $=($ amount dissociated $) /($ amount taken $)$

$
\begin{aligned}
& =\frac{1.82 \times 10^{-3}(0.05)}{0.05} \\
& =1.82 \times 10^{-3}
\end{aligned}
$

Question 6.53(b) Calculate the degree of ionization of 0.05M acetic acid if its pK_a value is 4.74. How is the degree of dissociation affected when its solution also contains

0.1M in HCl?

Answer :

Let the x amount of acetic acid is dissociated in this case

$\begin{array}{cccc}\mathrm{CH}_3 \mathrm{COOH} & \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} & & \\ \text {Initial conc. } & 0.05 & 0 & 0 \\ \text { after dissociation } & 0.05-x & 0.001+x & x\end{array}$

As the dissociation is very small.
So we can write $0.1+x \approx 0.1$ and $0.05-x \approx 0.05$

$\begin{gathered}K_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ =\frac{(0.1)(x)}{(0.05)}=2 x\end{gathered}$

So, the value of$x=\frac{\left(1.82 \times 10^{-4}\right)(0.05)}{0.1}$

Now the degree of dissociation = (amount dissociated) / (amount is taken)

$\begin{aligned} & =\frac{1.82 \times 10^{-4}(0.05)}{0.05} \\ & =1.82 \times 10^{-4}\end{aligned}$

Question 6.54 The ionization constant of dimethylamine is 5.4 × 10^-4 . Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?

Answer :

We have,

(Degree of ionization) $K_b=5.4 \times 10^{-4}$
Concentration of dimethylamine = 0.02 M

$\therefore \alpha=\sqrt{\frac{K_b}{C}}=\sqrt{\frac{5.4 \times 10^{-4}}{0.02}}$$\alpha=0.1643$

If we add 0.1 M of sodium hydroxide. It is a strong base so, it goes complete ionization

$\mathrm{NaOH} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{OH}^{-}$
(0.1 M) (0.1 M)

and also,

$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons\left(\mathrm{CH}_3\right)_2 \mathrm{NH}_2^{+}+\mathrm{OH}^{-}$
0.02- x x x

$\left[\mathrm{OH}^{-}\right]=x+0.1 \approx 0.1$ (since the dissociation is very small)

Therefore,

$\begin{aligned} & K_b=\frac{x(0.1)}{0.02} \\ & x=5.4 \times 10^{-4}\left(\frac{0.02}{0.1}\right) \\ & =0.0054\end{aligned}$

Hence in the presece of 0.1 M of sodium hydroxide , 0.54% of dimethylamine get dissociated.

Question 6.55(a) Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

Human muscle-fluid, 6.83

Answer :

We have pH 6.83

It is known that $pH=-\log \left[H^{+}\right]$

Therefore,
$6.83=-\log \left[H^{+}\right]$
By taking antilog on both sides we get,

$\left[H^{+}\right]=1.48 \times 10^{-7} M$

Question 6.55(b) Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

Human stomach fluid, 1.2

Answer :

We have $pH=1.2$

It is known that $pH=-\log \left[H^{+}\right]$

Therefore,
$1.2=-\log \left[H^{+}\right]$
By taking antilog on both sides we get,

$\left[H^{+}\right]=0.063 M$

Question 6.55(c) Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

Human blood, 7.38

Answer :

we have $pH=7.38$

It is known that pH=−log⁡[H+]

Therefore,
$7.38=-\log \left[H^{+}\right]$
By taking antilog on both sides we get,

$\left[H^{+}\right]=4.17 \times 10^{-8} M$

Question 6.55(d) Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

Human saliva, 6.4.

Answer :

we have $pH=6.4$

It is known that pH=−log⁡[H+]

Therefore,
$6.4=-\log \left[H^{+}\right]$
By taking antilog on both sides we get,

$\left[H^{+}\right]=3.98 \times 10^{-7} M$

Question 6.56 The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each

Answer :

We already know that pH can be calculated as- $-\log \left[H^{+}\right]$
to calculate the concentration of $\left[H^{+}\right]=\operatorname{antilog}\left(-pH\right)$

Thus, the hydrogen ion concentration of followings pH values are-

(i) pH of milk = 6.8
Since, pH=$-\log \left[H^{+}\right]$
6.8 = −log⁡[H+]
$\log \left[H^{+}\right]$ = -6.8

[H+] = anitlog(-6.8)

$=1.5 \times 10^{-7} M$

(ii) pH of black coffee = 5.0

Since,pH=$-\log \left[H^{+}\right]$

5.0 = $-\log \left[H^{+}\right]$

$\log \left[H^{+}\right]$= -5.0

$\left[H^{+}\right]$ = anitlog(-5.0)

$=10^{-5} M$

(iii) pH of tomato juice = 4.2

Since, pH=$-\log \left[\mathrm{H}^{+}\right]$

4.2 = $-\log \left[\mathrm{H}^{+}\right]$

$\begin{aligned} & \log \left[H^{+}\right]=-4.2 \\ & {\left[H^{+}\right]=\operatorname{anitlog}(-4.2)} \\ & =6.31 \times 10^{-5} \mathrm{M}\end{aligned}$

(iv) pH of lemon juice = 2.2

Since, pH=$-\log \left[H^{+}\right]$

2.2 = $-\log \left[H^{+}\right]$

$\begin{aligned} & 2.2=-\log \left[H^{+}\right] \\ & \log \left[H^{+}\right]=-2.2 \\ & {\left[H^{+}\right]=\operatorname{anitlog}(-2.2)} \\ & =6.31 \times 10^{-3} \mathrm{M}\end{aligned}$

(v) pH of egg white = 7.8

Since, pH=$-\log \left[\mathrm{H}^{+}\right]$

7.8 = $-\log \left[\mathrm{H}^{+}\right]$

$\begin{aligned} & \log \left[H^{+}\right]=-7.8 \\ & {\left[H^{+}\right]=\operatorname{anitlog}(-7.8)} \\ & =1.58 \times 10^{-8} M\end{aligned}$

Question 6.57 If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions.What is its pH?

Answer :

We have 0.562 g of potassium hydroxide ( KOH ). On dissolving in water gives 200 mL of solution.
Therefore, concentration of

$[K O H(a q)]=\frac{0.561 \times 1000}{200} g / L$

= 2.805 g/L

$=2.805 \times \frac{1}{56.11} M=0.05 M$
$\mathrm{KOH}(a q) \rightarrow \mathrm{K}^{+}(a q)+\mathrm{OH}^{-}(a q)$
It is a strong base. So, that it goes complete dissociation.

$\left[O H^{-}\right]=0.05 M=\left[K^{+}\right]$

It is known that,
$\begin{aligned} & K_w=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right]=\frac{K_w}{\left[\mathrm{OH}^{-}\right]} \\ & =\frac{10^{-14}}{0.05}=2 \times 10^{-13} \mathrm{M}\end{aligned}$

Therefore,$pH=-\log \left(2 \times 10^{-13}\right)=12.69$

Question 6.58 The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Answer :

By given abova data, we know the solubility of Sr(OH)₂ at 298 K = 19.23 g/L

So, concentration of [Sr(OH)₂]

= 19.23/121.63M (Molecular weight of Sr(OH)₂ = 121.63 u)

= 0.1581 M

$\mathrm{Sr}(\mathrm{OH})_2(\mathrm{aq}) \rightarrow \mathrm{Sr}^{2+}(\mathrm{aq})+2\left(\mathrm{OH}^{-}\right)(\mathrm{aq})$

$\therefore S r^{2+}=0.1581 M$
and the concentration of [OH⁻] =2×0.1581M=0.3162M
Now

It is known that,

$
\begin{aligned}
& K_w=\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right] \\
& {\left[\mathrm{H}^{+}\right]=\frac{10^{-14}}{0.3162}} \\
& =3.16 \times 10^{-14}
\end{aligned}
$

Therefore $pH=-\log \left(3.16 \times 10^{-14}\right)=13.5$

Question 6.60 The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Answer :

We have,
Concentration of cyanic acid = 0.1 M
pH=$-\log \left[H^{+}\right]=2.34$

Therefore, the concentration of [H+] = antilog (-2.34)

$=4.5 \times 10^{-3}$

It is known that,

$\begin{aligned} & {\left[H^{+}\right]=C . \alpha=4.5 \times 10^{-3}} \\ & \alpha=\frac{4.5 \times 10^{-3}}{0.1}=4.5 \times 10^{-2}\end{aligned}$

Then Ionization constant

$\left(K_a\right)=C \cdot \alpha^2$$\begin{aligned} & =(0.1)\left(4.5 \times 10^{-2}\right)^2 \\ & =2.02 \times 10^{-4}\end{aligned}$

Question 6.61 The ionization constant of nitrous acid is 4.5 × 10 ^-4 . Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Answer :

We have,
Ionization constant of nitrous acid =4.5×10⁻⁵
Concentration of sodium nitrite ( NaNO₂ ) = 0.04 M
Degree of hydrolysis can be calculated as;
$K_h=\frac{K_w}{K_a}=\frac{10^{-14}}{4.5 \times 10^{-4}}=0.22 \times 10^{-10}$
Sodium nitrite is a salt of sodium hydroxide (strong base) and the weak acid ( $\mathrm{HNO}_2$ )
$\mathrm{NO}_2^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{HNO}_2+\mathrm{OH}^{-}$
Suppose x moles of salt undergoes hydrolysis, then the concentration of-

$
\left[\mathrm{NO}_2^{-}\right]=0.04-x \approx 0.04
$

$\left[\mathrm{HNO}_2^{-}\right]=x$, and

$
\left[O H^{-}\right]=x
$

Therefore
$k_h=\frac{x^2}{0.04}=0.22 \times 10^{-10}$
from here we can calculate the value of x ;

$\begin{aligned} & \Rightarrow x=\sqrt{0.0088 \times 10^{-10}} \\ & =0.093 \times 10^{-5}=\left[\mathrm{OH}^{-}\right] \\ & \Rightarrow\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{K_w}{\left[\mathrm{OH}^{-}\right]}=\frac{10^{-14}}{0.093 \times 10^{-5}} \\ & =10.75 \times 10^{-9} \mathrm{M}\end{aligned}$

Therefore the degree of hydrolysis

Question 6.62 A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Answer :

Given,
pH = 3.44

We know that

pH=$-\log \left[H^{+}\right]$
By taking antilog on both sides we get,
[H+] = antilog (- 3.44)

$\therefore\left[H^{+}\right]=3.63 \times 10^{-4}$

pyridinium hydrochloride completely ionised.

Then K_h = (conc. of products)/ (conc, of reactants)
$=\frac{\left(3.63 \times 10^{-2}\right)^2}{0.02}$ (? Concentration is 0.02M)

$\Rightarrow K_h=6.58 \times 10^{-6} $

Now,$\begin{aligned} & K_h=K_w / K_a \\ & \Rightarrow K a=K_w / K_h\end{aligned}$

$\begin{aligned} & =10^{-14} / 6.58 \times 10^{-6} \\ & =1.51 \times 10^{-9} \text { (approx) }\end{aligned}$

Question 6.63 Predict if the solutions of the following salts are neutral, acidic or basic: $\mathrm{NaCl}, \mathrm{KBr}, \mathrm{NaCN}, \mathrm{NH}_4 \mathrm{NO}_3, \mathrm{NaNO}_2$ and KF

Answer :

Salts of strong acid and strong base are neutral in nature for example-

• NaCl(NaOH+HCl)
• KBr(KOH+HBr)

Salts of a strong base and weak acid are basic in nature for example-

• NaCN(HCN+NaOH)
• $\mathrm{NaNO}_2\left(\mathrm{HNO}_2+\mathrm{NaOH}\right)$
• KF(KOH+HF)

Salts of strong acid and a weak base are acidic in nature for example-

• $\mathrm{NH}_4 \mathrm{NO}_3\left(\mathrm{NH}_4 \mathrm{OH}+\mathrm{HNO}_2\right)$

Question 6.65 Ionic product of water at 310 K is 2.7 × 10 ^-14 . What is the pH of neutral water at this temperature?

Answer :

We have the ionic product of water at 310 K is $2.7 \times 10^{-14}$
It is known that,
ionic product $K_w=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]$

Since $\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]$, therefore $K_w=\left[\mathrm{H}^{+}\right]^2$

$\begin{aligned} & \Rightarrow K_w \text { at } 310 \mathrm{~K} \text { is } 2.7 \times 10^{-14} \\ & \therefore K_w=2.7 \times 10^{-14}=\left[H^{+}\right]^2\end{aligned}$
here we can calculate the value of [H+] concentration.

$
\left[H^{+}\right]=\sqrt{2.7 \times 10^{-14}}=1.64 \times 10^{-7}
$

Thus, $pH=-\log \left[H^{+}\right]$

$
\begin{aligned}
& =-\log \left(1.64 \times 10^{-7}\right) \\
& =6.78
\end{aligned}
$

Hence the pH of neutral water is 6.78

Question 6.66(a) Calculate the pH of the resultant mixtures:

10 mL of 0.2M Ca(OH)₂ + 25 mL of 0.1M HCl

Answer :

Given that,
Vol. of 0.2 M Ca(OH)₂ = 10 mL
Vol. of 0.1 M HCl = 25 mL

therefore, by using the formula,

$M\left(O H^{-}\right)=\frac{M_1 V_1(\text { base })-M_2 V_2(\text { acid })}{V_1+V_2}$
By substituting the value in these equations, we get;

$\Rightarrow \frac{(0.2 \times 2)-(0.1 \times 2)}{10+25}=\frac{1.5}{25}=0.06$

Now, pOH=$-\log \left[\mathrm{OH}^{-}\right]$

$=-\log (0.06)=1.221$

since pH+pOH=14
pH=14−pOH
= 14-1.221
= 12.78

Question 6.66(b) Calculate the pH of the resultant mixtures:

10 mL of 0.01M H₂SO₄ + 10 mL of 0.01M Ca(OH)₂

Answer :

In this case, both the solutions have the same number of moles of H⁺ and OH⁻ , therefore they both can get completely neutralised. Hence the pH = 7.0

Question 6.66(c) Calculate the pH of the resultant mixtures:

c) 10 mL of 0.1M H₂SO₄ + 10 mL of 0.1M KOH

Answer :

Given that,

Volume of 0.1 M KOH = 10 mL, and

Volume of 0.1 M H₂SO₄ = 10 mL

So, by using the formula of,

$M\left(H^{+}\right)=\frac{M_1 V_1(\text { acid })-M_2 V_2(\text { base })}{V_1+V_2}$
By putting the values we get,

$\Rightarrow \frac{2(0.1 \times 10)-0.1 \times 10}{10+10}=\frac{1}{20}=5 \times 10^{-2}$

Hence, $p H=-\log \left[H^{+}\right]=-\log \left(5 \times 10^{-2}\right)=1.30$

Question 6.67 Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table. Determine also the molarities of individual ions.

Answer :

Solubility product is the product of ionic concentrations in a saturated solution.
$K_{s p}=\left[A^{+}\right]\left[B^{-}\right]$
(i) silver chromate ( Ag₂CrO₄ )
Ionization of silver chromate

$\mathrm{Ag}_2 \mathrm{CrO}_4 \rightleftharpoons 2 \mathrm{Ag}^{+}+\mathrm{CrO}_4^{2-}$
Let " s " be the solubility of $\mathrm{Ag}_2 \mathrm{CrO}_4$
$\begin{aligned} & {\left[\mathrm{Ag}^{+}\right]=2 s} \\ & {\left[\mathrm{CrO}_4^{2-}\right]=s}\end{aligned}$
According to the table $K_{s p}$ of $\mathrm{Ag}_2 \mathrm{CrO}_4=1.1 \times 10^{-12}$

$\begin{aligned} & \Rightarrow 1.1 \times 10^{-12}=(2 s)^2 . s=1.1 \times 10^{-12}=2 s^3 \\ & s=\sqrt[3]{\frac{1.1 \times 10^{-12}}{4}} \\ & =0.65 \times 10^{-4}\end{aligned}$

(ii) Barium chromate ( BaCrO₄ )
Ionization of silver chromate

$\mathrm{BaCrO}_4 \rightleftharpoons \mathrm{Ba}^{2+}+\mathrm{CrO}_4^{2-}$
Let " s " be the solubility of BaCrO4
$\begin{aligned} & {\left[\mathrm{Ba}^{2+}\right]=s} \\ & {\left[\mathrm{CrO}_4^{2-}\right]=s}\end{aligned}$
According to the table $K_{s p}$ of $\mathrm{BaCrO}_4=1.2 \times 10^{-12}$

$\begin{aligned} & \Rightarrow 1.2 \times 10^{-10}=s . s=1.1 \times 10^{-10}=s^2 \\ & s=\sqrt{\frac{1.2 \times 10^{-10}}{1}} \\ & =1.09 \times 10^{-5}\end{aligned}$

(iii) Ferric hydroxide Fe(OH)₃)
Ionization of Ferric hydroxide

$\mathrm{Fe}(\mathrm{OH})_3 \rightleftharpoons \mathrm{Fe}^{3+}+3 \mathrm{OH}^{-}$
Let " s " be the solubility of Fe(OH)₃
$\begin{aligned} & {\left[F e^{3+}\right]=s} \\ & {\left[O H^{-}\right]=3 s}\end{aligned}$
According to the table Ksp of Fe(OH)3 = 1.0×10−38

$\begin{aligned} & \Rightarrow 1.0 \times 10^{-38}=s .(3 s)^3=1.0 \times 10^{-38}=27 s^4 \\ & s=\sqrt[4]{\frac{1.0 \times 10^{-38}}{27}} \\ & =1.39 \times 10^{-10}\end{aligned}$

(iv)

Lead chloride ( PbCl₂ )
Ionization of Lead chloride

$
\mathrm{PbCl}_2 \rightleftharpoons \mathrm{~Pb}^{2+}+2 \mathrm{Cl}^{-}
$

Let " $s$ " be the solubility of $\mathrm{PbCl}_2$

$
\begin{aligned}
& {\left[\mathrm{Pb}^{2+}\right]=s} \\
& {\left[\mathrm{Cl}^{-}\right]=2 s}
\end{aligned}
$

According to the table $K_{s p}$ of $\mathrm{PbCl}_2=1.6 \times 10^{-5}$

$\begin{aligned} & \Rightarrow 1.6 \times 10^{-5}=s \cdot(2 s)^2 \\ & =1.6 \times 10^{-5}=4 s^3 \\ & s=\sqrt[3]{\frac{1.6 \times 10^{-5}}{4}} \\ & =1.58 \times 10^{-2}\end{aligned}$

So molarity of $\mathrm{Pb}^{2+}=1.58 \times 10^{-2} \mathrm{M}$ and molarity of $\mathrm{Cl}^{-}=3.16 \times 10^{-2} \mathrm{M}$

Question 6.68 The solubility product constant of Ag₂CrO₄ and AgBr is 1.1 × 10 -12 and 5.0 × 10 -13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Answer :

silver chromate $\left(\mathrm{Ag}_2 \mathrm{CrO}_4\right)$
Ionization of silver chromate

$\mathrm{Ag}_2 \mathrm{CrO}_4 \rightleftharpoons 2 \mathrm{Ag}^{+}+\mathrm{CrO}_4^{2-}$
Let " s " be the solubility of $\mathrm{Ag}_2 \mathrm{CrO}_4$

$\begin{aligned} & {\left[\mathrm{Ag}^{+}\right]=2 s} \\ & {\left[\mathrm{CrO}_4^{2-}\right]=s} \\ & K_{s p} \text { of } \mathrm{Ag}_2 \mathrm{CrO}_4=1.1 \times 10^{-12} \\ & \Rightarrow 1.1 \times 10^{-12} \\ & =(2 s)^2 \cdot s=1.1 \times 10^{-12}=2 s^3 \\ & s=\sqrt[3]{\frac{1.1 \times 10^{-12}}{4}} \\ & =0.65 \times 10^{-4}\end{aligned}$

Ionization of Silver bromide ( AgBr )

$\begin{aligned} & A g B r \rightleftharpoons A g^{+}+B r^{-} \\ & K_{s p} \text { of } A g B r=5 \times 10^{-13} \\ & {\left[\mathrm{Ag}^{+}\right]=s^{\prime}} \\ & {\left[\mathrm{Br}^{-}\right]=s^{\prime}} \\ & \Rightarrow 5 \times 10^{-13}=s^{\prime} . s^{\prime} \\ & =5 \times 10^{-13}=s^{\prime 2} \\ & s^{\prime}=\sqrt{\frac{5 \times 10^{-13}}{1}}=\sqrt{0.5 \times 10^{-12}} \\ & =7.07 \times 10^{-7}\end{aligned}$

Now, the ratio of solubilities

$\begin{aligned} & \Rightarrow \frac{s}{s^{\prime}}=\frac{6.5 \times 10^{-5}}{1.1 \times 10^{-12}} \\ & =9.91\end{aligned}$

Question 6.69 Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Ksp = 7.4 × 10^-8 ).

Answer :

We have,
solubility product $\left(K_{s p}\right)$ of cupric iodate $=7.4 \times 10^{-8}$

When equal volumes of sodium iodate and cupric chlorate are mixed together the molar concentration of both the solution becomes half (= 0.001)

Ionization of cupric iodate is;

$\underset{0.001 \mathrm{M}}{\mathrm{Cu}\left(\mathrm{IO}_3\right)_2} \rightleftharpoons \mathrm{Cu}^{2+}+\underset{0.001 \mathrm{M}}{2 \mathrm{IO}_3^{-}}$

So, Ksp can be calculated as;

$\begin{aligned} & K_{s p}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{IO}_3^{-}\right]^2 \\ & =(0.001)^3=10^{-9}\end{aligned}$

Sinc eionic product is less than the Ksp so no precipitation occurs.

Question 6.70 The ionization constant of benzoic acid is 6.46 × 10^-5 and Ksp for silver benzoate is 2.5 × 10^-13 . How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Answer :

Suppose S is the molar solubility of silver benzoate in water, then

$C_6 \mathrm{H}_5 \mathrm{COOAg}_s \rightleftharpoons C_6 \mathrm{H}_5 \mathrm{COO}_{a q}^{-}+\mathrm{Ag}_{a q}^{+}$

so S

$=\sqrt{2.5 \times 10^{-13}}=5.0 \times 10^{-7} \mathrm{M}$

If the solubility of salt of weak acid of ionization constant Ka is S , then Ksp,Ka and S′ are related to each other at pH=3.19.

So$\left[H^{+}\right]=6.46 \times 10^{-4} M(\because \mathrm{pH}=3.19)$

$K_{s p}={S^{\prime}}^2\left[\frac{K_a}{K_a+\left[H^{+}\right]}\right]$

$S^{\prime}=\left\{\frac{2.5 \times 10^{-13}}{\left[\frac{6.46 \times 10^{-5}}{6.46 \times 10^{-5}+6.46 \times 10^{-4}}\right]}\right\}^{\frac{1}{2}}$

$S^{\prime}=\left\{\frac{2.5 \times 10^{-13} \times 7.106 \times 10^{-4}}{6.46 \times 10^{-5}}\right\}^{\frac{1}{2}}$

$=\left(2.75 \times 10^{-12}\right)^{\frac{1}{2}}$

$=1.658 \times 10^{-6} M$

So the ratio of

$\frac{S I}{S}=\frac{1.658 \times 10^{-6}}{5.0 \times 10^{-7}}=3.32$
Silver benzoate is 3.32 times more soluble in buffer of $p H=3.19$ than in pure water.

Question 6.71 What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10^-18 ).

Answer :

We have,
The solubility product of the $F e S=6.3 \times 10^{-18}$

Equals number of moles of ferrous sulphate and sodium sulphide are mixed in an equal volume.

Let s be the concentration of ferrous sulphate and sodium sulphide. On mixing the equimolar solution, the volume of the concentration becomes half.
So, $\left[\mathrm{FeSO}_4\right]=\left[\mathrm{Na}_2 \mathrm{~S}\right]=\frac{s}{2} M$

The ionisation of ferrous sulphide;
$F e S \rightleftharpoons F e^{2+}+S^{2-}$
Therefore, for no precipitation, ionic product = solubility product
$K_{s p}=\left(\frac{s}{2}\right)\left(\frac{s}{2}\right) 6.3 \times 10^{-18}=\frac{s^2}{4}$
By solving the above equation, we get
$s=5.02 \times 10^{-9}$

The maximum concentration of both the solution is$s=5.02 \times 10^{-9}$M

Question 6.72 What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, K sp is 9.1 × 10 ^-6 )

Answer :

We have,
The solubility product of calcium sulphate is 9.1×10⁻⁶ .
given mass of calcium sulphate = 1g

Ionization of calcium sulphate;

$\mathrm{CaSO}_4 \rightleftharpoons \mathrm{Ca}^{2+}+\mathrm{SO}_4^{2-}$
Therefore,$\mathrm{K}_{s p}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]$

Let the solubility of calcium sulphate be s .
Then,$\begin{aligned} & K_{s p}=s^2 9.1 \times 10^{-6} \\ & =s^2 s=\sqrt{9.1 \times 10^{-6}} \\ & =3.02 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\end{aligned}$

Thus,
mass/ (mol. wt) × volume =3.02×10⁻³ Molarity
mass$=3.02 \times 136 \times 10^{-3}=0.41 \mathrm{~g}$

So, that to dissolve 1 g of calcium sulphate we need $=1 / 0.41 L=2.44 L$of water.

Question 6.73 The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10 ^-19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO₄, MnCl₂, ZnCl₂ and CdCl₂. in which of these solutions precipitation will take place?

Answer :

We have,
the concentration of [S²⁻]=1×10⁻¹⁹ and the volume of the solution containing sulphur ion = 10 mL.
Volume of metal salts solution added = 5mL
Before mixing,

$\begin{aligned} & {\left[S^{2-}\right]=1 \times 10^{-19} \mathrm{M}} \\ & {\left[M^{2+}\right]=0.04 M}\end{aligned}$

After mixing,

Volume = 15 mL

So, the concentration of$\left[S^{2-}\right]=1 \times 10^{-19}$

$=6.67 \times 10^{-20} M$

$\begin{aligned} & \text { concentration of }\left[M^{2+}\right]=0.04 M \times \frac{5}{15} \\ & =1.33 \times 10^{-2}\end{aligned}$

$\begin{aligned} & \text { So, the ionic product }=\left[M^{2+}\right]\left[S^{2-}\right] \\ & =\left(6.67 \times 10^{-20}\right)\left(1.33 \times 10^{-2}\right) \\ & =8.87 \times 10^{-22}\end{aligned}$

For the precipitation of the solution, the ionic product should be greater than the corresponding Ksp values.

$\mathrm{K}_{s p}$ of FeS, MnS, ZnS, CdS are $6.3 \times 10^{-18}, 2.5 \times 10^{-13}, 1.6 \times 10^{-24}$ and $8 \times 10^{-27}$ respectively.

Hence precipitation will take place in CdCl₂ and ZnCl₂ metal salts.