Question

An acid $HA$ ionizes as, $HA\rightleftharpoons H^{+}+A^{-}$. The pH of $1.0$M solution is $5$. Its dissociation constant would be:

• A. $1\times {10}^{-10}$
• B. $5$
• C. $5\times {10}^{-8}$
• D. $1\times {10}^{-5}$

Answer 1

Answer: (A)

$1\times {10}^{-10}$

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

Initial $1m00$

equilibrium $1-xxx$

given $\{pH=5\}$

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

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

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

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

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

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