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Ionization Of Acids And Bases MCQ - Practice Questions with Answers

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

Syllabus

Quick Facts

8 Questions around this concept.

Solve by difficulty

DIFFICULT

The ionization constant of ammonium hydroxide is $1.77\times 10^{-5}$ at 298K. Hydrolysis constant of ammonium chloride is:

$6.50\times 10^{-12}$

$5.65\times 10^{-13}$

$5.65\times 10^{-12}$

$5.65\times 10^{-10}$

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Dielectric constant of ordinary water is ______ than heavy water.

Less

Equal

Can not be predicted.

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Concepts Covered - 2

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.
$\mathrm{HA(aq)\: +\: H_{2}O(l)\: \rightleftharpoons \: H_{3}O^{+}\: +\: A^{-}(aq)}$

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

𝛂 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, 𝛂 is small, thus 1 - 𝛂 = 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:
$\mathrm{B(aq)\: +\: H_{2}O(l)\: \rightleftharpoons \: HB^{+}\: +\: OH^{-}(aq)}$

Thus,
$\mathrm{BOH\: \rightleftharpoons \: B^{+}\: +\: OH^{-}}$
$\mathrm{K_{b}\: =\: \frac{c\alpha ^{2}}{(1-\alpha )}}$
Thus, if 1 - 𝛂 = 1, then
$\mathrm{K_{b}\: =\: c\alpha ^{2}\: or\: \alpha \: =\: \sqrt{\left ( \frac{K_{b}}{c} \right )}}$

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.