Question

If $K_a$ for an acid, $HA$ is $1\times {10}^{-6}$, then $K_b$ for $A^{-}$ would be:

• A. ${10}^{-9}$
• B. ${10}^{-5}$
• C. ${10}^{-8}$
• D. ${10}^{-3}$

Answer 1

The correct option is C ${10}^{-8}$

The acid dissociation equilibrium is as given below:

$HA\rightleftharpoons H^{+}+A^{-}$ $K_a=1\times {10}^{-6}$

The base dissociation equilibrium is as follows:

$A^{-}+H_{2}O\rightleftharpoons HA+O{H}^{-}$

$K_b=\frac{K_w}{K_a}=\frac{{10}^{-14}}{{10}^{-6}}={10}^{-8}$

Hence, the correct option is $\text{C}$