Question

$p{K}_a$ for the acid $HA$ is $6$. The value of $K$ for the reaction $A^{-}+H_3{O}^{+}\rightleftharpoons HA+H_{2}O$ is:

• A. $1\times {10}^{-6}$
• B. $1\times {10}^8$
• C. $1\times {10}^{-8}$
• D. $1\times {10}^6$

Answer 1

The correct option is D $1\times {10}^6$

The acid dissociation equilibrium is as shown below
$HA+H_{2}O\rightleftharpoons H_3{O}^{+}+A^{-}$
The acid dissociation constant is
$K_a={10}^{-6}$
The reaction $HA+H_{2}O\rightleftharpoons H_3{O}^{+}+A^{-}$ is the reverse of the reaction for the acid dissociation equilibrium.
Hence, $K=\frac{1}{K_a}\frac{1}{{10}^{-6}}={10}^6$