Name: Common Ion Effect on Solubility Product
Uploaded: 2019-07-19
Duration: 18 min 45 s
Description: The presence of a common ion leads to a change in the extent of a reaction, according to Le-Chatelier's principle. Watch this video to learn how the common i...

Question

Calculate the

i) hydrolysis constant,

ii) degree of hydrolysis and

iii) $pH$ of $0.05 M$ sodium carbonate solution $({pK}_{a}$ for ${HCO}_{3}^{-}$ is $10.26$ ).

Accepted Answer

(i) An equilibrium constant for a hydrolysis reaction is called hydrolysis constant. It is denoted by $K_{h}$ .

Hydrolysis constant for a slat of weak acid and strong base(sodium carbonate) is :

$K_{h} = \frac{K_{w}}{K_{a}}$

We know that, $K_{w} = 1.0 \times 10^{- 14}$

Given, ${pK}_{a} = 10.26$

${pK}_{a} = - \log K_{a}$

$K_{a} = antilog (- {pK}_{a}) = antilog (- 10.26) = 5.49 \times 10^{- 11}$

$K_{h} = \frac{1.0 \times 10^{- 14}}{5.49 \times 10^{- 11}} = 1.82 \times 10^{- 4}$

(ii) Degree of hydrolysis :

We also know that the degree of hydrolysis for a slat of weak acid and strong base(sodium carbonate) is:

$α = \sqrt{\frac{K_{w}}{K_{a} . C}} α = \sqrt{\frac{1.0 \times 10^{- 14}}{5.49 \times 10^{- 11} \times 0.05}} α = \sqrt{3.642 \times 10^{- 3}}$

Therefore, degree of hydrolysis, $α = 6.034 \times 10^{- 2}$

(iii) pH :

The relation between $pH, {pK}_{a} and C$ is :

$pH = 7 + \frac{1}{2} {pK}_{a} + \frac{1}{2} \log C$

$pH = 7 + \frac{1}{2} \times 10.26 + \frac{1}{2} \log 0.05$

$pH = 7 + 5.13 - 0.65$

$pH = 11.48$

Tamil Nadu Board Solutions for Chapter: Ionic Equilibrium, Exercise 10: Evaluate yourself - 10

Tamil Nadu Board Chemistry Solutions for Exercise - Tamil Nadu Board Solutions for Chapter: Ionic Equilibrium, Exercise 10: Evaluate yourself - 10

Questions from Tamil Nadu Board Solutions for Chapter: Ionic Equilibrium, Exercise 10: Evaluate yourself - 10 with Hints & Solutions

Learn and Prepare for any exam you want !