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Common ion effect MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Syllabus

Quick Facts

pH of weak acid + strong acid is considered one the most difficult concept.
9 Questions around this concept.

Solve by difficulty

EASY

Which one of the following statements is not true?

The conjugate base of $H_{2}PO{_{4}}^{-}$ is $HPO{_{4}}^{2-}$

$pH+pOH= 14\: for \: all \: aqueous\: solutions.$

$The\: pH\: of\: 1\times 10^{-8}M\: H\! C\! L \: is \: 8$

96,500 coulombs of electricity when passed through a $CuSO_4$ solution deposits 1 gram equivalent of copper at the cathode.

View Solution

Concepts Covered - 2

Common ion effect

The value of degree of dissociation for a weak electrolyte is decreased by the addition of a strong electrolyte having common ion. As a result of this effect, the concentration of the uncommon ion of the weak electrolyte decreases.
For example:

$\mathrm{NH_{4}OH\: \leftrightarrow \: NH_{4}^{+}\: +\: OH^{-}}$
weak electrolyte

$\mathrm{NH_{4}Cl\: \leftrightarrow \: NH_{4}^{+}\: +\: Cl^{-}}$
Common ion
Here $\alpha$ for NH₄OH will be decreased by NH₄Cl
$\mathrm{CH_{3}COOH\: \leftrightarrow \: CH_{3}COO^{-}\: +\: H^{+}}$
$\mathrm{CH_{3}COONa\: \leftrightarrow \: CH_{3}COO^{-}\: +\: Na^{+}}$
Here $\alpha$ for CH₃COOH will be decreased by CH₃COONa

Applications of Common ion effect

Solubility of a partially soluble salt decreases due to common ion effect. For example, presence of AgNO₃or KCl decreases solubility of AgCl in water.
Salting out of soap by addition of NaCl.
Purification of NaCl by passing HCl gas.

Isohydric Solution: These are the solutions having same concentration of common ions.

pH of weak acid + strong acid

The pH of a mixture of a weak acid and a strong acid can be understood using the following example.

Weak acid: H₂S(0.1M)
Strong acid: HCl(0.3M)
The K_a value for this mixture = 1.2 x 10^-20

The chemical equation for H₂S is given as follows:

$\mathrm{H_{2}S\: \rightleftharpoons \: 2H^{+}\: +\: S^{2-}}$
Initial: c 0 0
Equil: c - c? c? c?

Thus, the equilibrium constant K_a is given as follows:

$\\\mathrm{K_{a}\: =\: \frac{[c\alpha ]^{2}.c\alpha }{c(1-\alpha )}\: =\: \frac{[H^{+}]^{2}.c\alpha }{c(1-\alpha )}}\\\\\mathrm{1.2\, \,x\, 10^{-20}\: = \frac{[c\alpha \: +\: 0.3].c\alpha }{c(1-\alpha )}\: =\:(0.3)^{2}\: =\: 0.09\alpha}\\\\\mathrm{\mathbf{Thus},\: \alpha \: =\: \frac{1.2\: x\: 10^{-20}}{0.09}\: =\: 13.3\: x\: 10^{-20}}$

Again, the chemical equation for HCl is given as follows:
$\mathrm{HCl\: \rightarrow \: H^{+}\: +\: Cl^{-}}$
Initial: 0.3 0 0
Equil: 0 0.3 0.3

Thus, total [H⁺] for the mixture = c? + 0.3
[H⁺] = 3 x 10^-1 (? is very small for weak acids)

Now, the pH for the mixture is given as follows:
$\\\mathrm{pH\: =\: -log[H^{+}]\: =\: -log[3\: x\: 10^{-1}]\: =\: -log\, 3\: +\: 1\: =\: -0.47\: +\: 1}\\\\\mathrm{pH\: =\: 0.53}$