Question

$K_a$ for butyric acid is $2.0\times {10}^{-5}$. Calculate pH and hydroxyl ion concentration in $0.2M$ aqueous solution of sodium butyrate.

• A. $pH=5;[OH{]}^{-}={10}^{-9}M$
• B. $pH=5;[OH{]}^{-}={10}^{-5}M$
• C. $pH=9;[OH{]}^{-}={10}^{-9}M$
• D. $pH=9;[OH{]}^{-}={10}^{-5}M$

Answer 1

The correct option is D $pH=9;[OH{]}^{-}={10}^{-5}M$

$NaBu+H_{2}O\rightleftharpoons NaOH+BuH$
$\text{Initial conc.}100$
$\text{Conc. after hydrolysis}1-\alpha \alpha \alpha$
From hydrolysis of salt, we know,
$\left[O{H}^{-}\right]=C\alpha =C.\sqrt{\left(\frac{K_h}{C}\right)}=\sqrt{(K_h.C)}$
$=\sqrt{[\left(\frac{K_w}{K_a}\right).C]},where,K_h=\frac{K_w}{K_a}$
$\left[O{H}^{-}\right]=\sqrt{\frac{{10}^{-14}\times 0.2}{2\times {10}^{-5}}}=\sqrt{{10}^{-10}}={10}^{-5}$
$pOH=5$
$pH+pOH=14$
So $pH=14-5=9$