Chemistry - Ionic Equilibrium Question with Solution | TestHub

Explain the question : Calculate K for weak acid-strong base reaction.

Concept : This question is based on relationship between $\mathrm{K},{\mathrm{K}}_{\mathrm{a}}$, and ${\mathrm{K}}_{\mathrm{w}}$.

Solution : The reaction of a weak acid ( HA ) with a strong base (like NaOH ) is an acid base neutralization reaction: $\mathrm{HA}+{\mathrm{OH}}^{-}\rightleftharpoons {\mathrm{A}}^{-}+{\mathrm{H}}_2\mathrm{O}$. The equilibrium constant, K , for this reaction can be obtained by combining the acid dissociation of the weak acid and the auto ionization of water:

1. HA ⇌ H⁺ + A⁰ (equilibrium constant is K_a )

2. H⁺ + OH^- ⇌ H₂O (equilibrium constant is 1/K_w )

Summing these two reactions gives the neutralization reaction, so the overall equilibrium constant K is the product of the individual equilibrium constants : $\mathrm{K}={\mathrm{K}}_{\mathrm{a}}\times \frac{1}{{\mathrm{K}}_w}$ .

Using the given value K_a = 1 × 10^-4 and the standard value for the ionic product of water at 25°C, K_w = 1 × 10^-14 , we calculate the equilibrium constant

$\mathrm{K}=\frac{1\times 10^{-4}}{1\times 10^{-14}}=1\times 10^{10}$

Final Answer: Option (C)