Ka and Kb Relationship MCQ - Practice Questions & Answers

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What is the PH of a 1M CH₃COONa? Ka of acetic acid=1.8x10^-5,K_w=10^-14 mol²/l²

2.4

3.6

4.8

9.4

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The pH of 1N aqueous solutions of HCl, CH₃COOH and HCOOH follows the order

HCl > HCOOH > CH₃COOH

HCl = HCOOH > CH₃COOH

CH₃COOH > HCOOH > HCl

CH₃COOH = HCOOH > HCl

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Concepts Covered - 1

Relation Between Ka, Kb and Kw for Conjugate Pair

The relation between K_a,K_b and K_w can be understand by the following reaction.
$\mathrm{NH_{3}\: +\: H_{2}O\: \rightleftharpoons \: NH_{4}^{+}\: +\: OH^{-}}$
Base Conjugate acid

The equilibrium constant K_b for base NH₃ is given as:
$\mathrm{K_{b}\: =\: \frac{[NH_{4}^{+}][OH^{-}]}{NH_{3}}}$
Again, the reaction is:
$\mathrm{NH_{4}^{+}\: +\: H_{2}O\: \rightleftharpoons \: NH_{3}\: +\: H_{3}O^{+}}$

The equilibrium constant K_a for conjugate acid NH₄⁺ is given as:
$\mathrm{K_{a}\: =\: \frac{[NH_{3}][H_{3}O^{+}]}{NH_{4}^{+}}}$

Thus, $\mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{b}}=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{w}}$
Thus, $\mathrm{K}_{\mathrm{a}} \propto 1 / \mathrm{K}_{\mathrm{b}}$ at a given temperature

Therefore, if K_a of acid increases, then K_b of the conjugate base decreases. In other words, the conjugate base of strong acid is a weak base and vice-versa.

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