JEE Main Deleted Syllabus 2025: Reduced Chapters & Topics PCM

Ionization Of Acids And Bases - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

14 Questions around this concept.

Concepts Covered - 2

Ionisation of weak acids and Ostwald’s dilution law

pH of Weak Acids
Weak acids are those acids which dissociate partially in solutions. For example:

8 M HA (K_a = 2 x 10^-8)
The chemical equation for the dissociation of weak acid HA is as follows:
$\mathrm{HA\: \rightleftharpoons \: H^{+}\: +\: A^{-}}$
Initial: 8M 0 0
Equil: 8 - 8? 8? 8?

The equilibrium constant K_a for the weak acid is given as follows:
$\\\mathrm{K_{a}\: =\: \frac{[H^{+}][A^{-}]}{[HA]}\: =\: \frac{8\alpha \, .\, 8\alpha }{8(1-\alpha )}\: =\: \frac{8\alpha ^{2}}{1-\alpha }}\\\\\mathrm{K_{a}\: =\: 8\alpha ^{2}\: \: \: \: \: \: \: (as\: (1-\alpha) \: \approx \: 1)}\\\\\mathrm{Thus,\: \alpha \: =\: \sqrt{\frac{K_{a}}{8}}\: =\: \sqrt{\frac{2\, x\, 10^{-8}}{8}}\: =\: \sqrt{\frac{10^{-8}}{4}}\: =\: 0.5\, x\, 10^{-4}}\\\\\mathrm{[H^{+}]\: =\: 8\,\, x\,\, 0.5\,\, x\,\, 10^{-4}\: =\: 4\: x\: 10^{-4}}\\\\\mathrm{pH\: =\: -log_{10}4\: +\: 4}\\\\\mathrm{pH\: =\: -0.60\: +\: 4\: =\: 3.4}$
Thus, pH of this given acid = 3.4
0.002N CH₃COOH(? = 0.02)
The chemical equation for the dissociation of CH₃COOH is as follows:
$\mathrm{CH_{3}COOH\: \rightleftharpoons \: CH_{3}COO^{-}\: +\: H^{+}}$
Initial: c 0 0
Equil: c - c? c? c?

The equilibrium constant K_a for the weak acid is given as follows:
$\\\mathrm{K_{a}\: =\: \frac{[CH_{3}COOH^{-}][H^{+}]}{[CH_{3}COOH]}\: =\: \frac{c\alpha \, .\, c\alpha }{c(1-\alpha )}\: =\: \frac{c\alpha ^{2}}{1-\alpha }}\\\\\mathrm{K_{a}\: =\: c\alpha ^{2}\: \: \: \: \: \: \: (as\: (1-\alpha) \: \approx \: 1)}\\\\\mathrm{Now,\: as\: we\: have\: given:}\\\mathrm{c\: =\: 0.002N\: or\: 0.002M\: \: \: \: \: \: (Normality\: =\: Molarity,\: as\: n\: factor\: =\: 1)}\\\mathrm{\alpha \: =\:\frac{2}{100}\: =\: 0.02}\\\\\mathrm{Thus,\: [H^{+}]\: =\: 0.002\,\, x\,\, 0.02\: =\: 4\: x\: 10^{-5}}\\\\\mathrm{pH\: =\: -log_{10}(4\: x\: 10^{-5})}\\\\\mathrm{pH\: =\: 5\: -\:log 4\: =\: 4.4}$
Thus, pH of acetic acid = 4.4

Ostwald's Dilution Law

This is an application of law of mass action for weak electrolyte dissociation equilibria. Consider ionisation of a weak electrolyte say a monoprotic acid, acid HA.
$\mathrm{HA(aq)\:\: \rightleftharpoons \: H^{+}\: (aq)+\: A^{-}(aq)}$

Thus,
$\mathrm{HA\: \rightleftharpoons \: H^{+}\: +\: A^{-}}$
Moles before dissociation 1 0 0
Moles after dissociation 1 - ? ? ?

? is the degree of dissociation of weak acid HA and c is the concentration.

Thus, according to equilibrium constant equation, we have:
$\\\mathrm{K_{a}\: =\: \frac{[H^{+}][A^{-}]}{[HA]}\: =\: \frac{c\alpha .c\alpha }{c(1-\alpha )}}\\\\\mathrm{K_{a}\: =\: \frac{c\alpha ^{2}}{(1-\alpha )}}$
For weak electrolytes, ? is small, thus 1 - ? = 1
$\mathrm{K_{a}\: =\: c\alpha ^{2}\: or\: \alpha \: =\: \sqrt{\left ( \frac{K_{a}}{c} \right )}}$

Similar expression can be made for a weak base as BOH:
$\mathrm{BOH(aq)\: \: \rightleftharpoons \: B^{+}(aq)\: +\: OH^{-}(aq)}$

Similarly for the base BOH, the expression of K_b can be written as
$\mathrm{BOH\: \rightleftharpoons \: B^{+}\: +\: OH^{-}}$
$\mathrm{K_{b}\: =\: \frac{c\alpha ^{2}}{(1-\alpha )}}$
Thus, if 1 - ? = 1, then
$\mathrm{K_{b}\: =\: c\alpha ^{2}\: or\: \alpha \: =\: \sqrt{\left ( \frac{K_{b}}{c} \right )}}$

From the expression for Ka or K_b it is evident that

(1) As the value of concentration decreases, the degree of dissociation increases

(2) As the value of concentration increases, the degree of dissociation decreases

This is called as Ostwald's dilution law for weak electrolytes

Ionization Constant of Water / Ionic Product of Water

The ionisation of water occurs as follows:
$\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$

The equilibrium constant here is defined in a different way, and is called as ionic product K_wof water and is given by:
$\mathrm{K}_{\mathrm{w} }=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]$

At 25⁰C, K_w = 1.0 x 10^-14

Experimentally it has been seen that the K_w value changes on increasing or decreasing the temperature. At 63⁰C, K_w = 10^-13 and at 11⁰C, K_w = 0.3 x 10^-¹⁴

$\mathrm{since\: pure\: water\: is\: neutral,\: \left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]=\sqrt{\mathrm{K}_{\mathrm{W}}}=10^{-7} \mathrm{M} \: at\: 25^{\circ} \mathrm{C}}$

If a strong acid is added to it, [H⁺] increases and hence [OH^-] < 10^-7M at 25⁰C and solution is said to be acidic.
If a strong base is added to it, [OH^-] increases and hence [H⁺] must decrease in order to keep K_w constant. Now [OH^-] > 10^-7M and solution is basic (or alkaline).

Temperature dependence of Equilibrium Constant: Vant Hoff's Equation

$\mathrm{ \ell n \left [ \frac{K_{T_2}}{K_{T_1}} \right ]=\frac{\Delta H}{R}\left [ \frac{1}{T_1}-\frac{1}{T_2} \right ]}$

$\Rightarrow\mathrm{ 2.303\ log_{10} \left [ \frac{K_{T_2}}{K_{T_1}} \right ]=\frac{\Delta H}{R}\left [ \frac{1}{T_1}-\frac{1}{T_2} \right ]}$

Using the above equation, the value of Keq at any unknown temperature can be calculated if the Keq value at a particular temperature and $\mathrm{\Delta H}$ is known.

Conversely, the above equation can also be used to calculate the value of $\mathrm{\Delta H}$ if the values of Keq at two different temperatures are known.