Question

A certain weak acid has $K_a=1.0\times {10}^{-4}$. The equilibrium constant for tis reaction with a strong base is $K_{eq}={10}^{5X}$. Value of X is________.

Answer 1

Given $\underset{weak}{HA}+\underset{strong}{BOH}\rightleftharpoons BA+H_{2}O$
or $HA+B^{+}+{OH}^{-}\rightleftharpoons B^{+}+A^{-}+H_{2}O$
or $HA+{OH}^{-}\rightleftharpoons A^{-}+H_{2}O$
$\therefore$ $K_{eq}=\frac{\left[A^{-}\right]}{\left[HA\right]\left[{OH}^{-}\right]}.......\left(i\right)$
Also for weak acid $HA:HA\rightleftharpoons H^{+}+A^{-}$
$K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}.......\left(ii\right)$
By eqs. $\left(ii\right)$ and $\left(i\right)$ $\left(K_a/K_{eq}\right)=K_w$
$K_{eq}=\frac{K_a}{K_w}=\frac{{10}^{-4}}{{10}^{-14}}={10}^{10}$