Question

A certain weak acid has $K_a={10}^{-5}$. If the equilibrium constant for its reaction with a strong base is represented by $y\times {10}^{10}$, then find the value of $y$.

Answer 1

The acid dissociation equilibrium is as shown blow.
$HA\rightleftharpoons H^{+}+A^{-}$ $K_a=1\times {10}^{-5}$. ......(1)
The equilibrium for the reaction of acid with base is as given below.
$HA+B^{+}+O{H}^{-}\rightleftharpoons A^{-}+B^{+}+H_{2}O$ .........(2)
The equilibrium for dissociation of water is $H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$ $K_w=1\times {10}^{-14}$ ......(3)

Let, $K$ be the equilibrium constant for reaction (2)
When reaction 3 is subtracted from reaction 1, reaction 2 is obtained.
Hence, $K=\frac{K_a}{K_w}$
Substitute values in the above expression, we get
$1\times {10}^y=\frac{1\times {10}^{-5}}{1\times {10}^{-14}}$
Hence, $y=9$