Question

$K_a$ for an acid, $HA$ is $1\times {10}^{-6}$ and $K_b$ for $A^{-}$ is ${10}^{-x}$. Here, the value of $x$ is:

Answer 1

The acid dissociation equilibrium is as given below:

$HA\rightleftharpoons H^{+}+A^{-};$ $K_a$

The base dissociation equilibrium is as given below:

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

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