Question

An acid $HA$ has dissociated as the following manner
$HA\rightleftharpoons H^{+}+A^{-}$
It has concentration $1$ M and $pH=5$ then find out dissociation constant.

• A. $1\times {10}^{-10}$
• B. $1\times {10}^{-6}$
• C. $5\times {10}^{-5}$
• D. $5$

Answer 1

The correct option is A $1\times {10}^{-10}$

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

Initial equilibrium $1m00$

$1-x$ $x$ $x$

given $=pH=5$

$5=-\log \left[H^{+}\right]$

$\left[H^{+}\right]={10}^{-5}$

So $\{x={10}^{-5}\}$

$K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{HA}$

$K_a=\frac{\left({10}^{-5}\right)\left({10}^{-5}\right)}{1-{10}^{-5}}$

$K_a=1\times {10}^{-10}$ $\left\{\right(1-{10}^{-5})\approx 1\}$