Question

The degree of hydrolysis and pH of $0.1M$ sodium acetate solution are:
Hydrolysis constant of sodium acetate is $5.6\times {10}^{-10}$.

• A. $h$ = $7.5\times {10}^{-7}M$, $pH$ = $7.9$
• B. $h$ = $8.5\times {10}^{-6}$, $pH$ = $6.9$
• C. $h$ = $8.5\times {10}^{-7}$, $pH$ = $7.9$
• D. $h$ = $7.5\times {10}^{-6}$, $pH$ = $7.9$

Answer 1

The correct option is D

$h$ = $7.5\times {10}^{-6}$, $pH$ = $7.9$

When we have salts of strong base and weak acid,

$h$ = $\sqrt{\frac{K_h}{c}}$

= $7.5\times {10}^{-6}$.
Note: you don't actually have to calculate this, just be careful about the power of 10 and the fact that $\sqrt{5}6$ lies between 7 and 8. Look at the options and you can figure out what h will be.

The hydrolysis of Sodium acetate happens as:

$C{H}_{3}CO{O}^{-}+H_{2}O\rightleftharpoons C{H}_{3}COOH+O{H}^{-}$

$c(1-h)$ $ch$ $ch$

$\left[O{H}^{-}\right]$ = $\left[C{H}_{3}COOH\right]$ = $ch$ = $7.5\times {10}^{-7}M$

$\left[C{H}_{3}CO{O}^{-}\right]$ = $c(1-h)$ = $(1-7.5\times {10}^{-6})\times 0.1mol{L}^{-1}\approx 1M$

$pH$ = $-Log\frac{Kw}{\left[O{H}^{-}\right]}$ = $8-log1.3$ = $7.9$
Note: Again, you don't have to know the value of log 1.3, just that it is going to be <1 and looking at the options, you can figure out the answer.