Question

Ionization constant of acetic acid and ionic product of water at $25^{\circ }C$ are respectively $1.75\times 10^{-5}$ and $1\times 10^{-4}$ respectively. Calculate the hydrolysis constant of sodium acetate and its degree of hydrolysis in $10^{-3}M$ solution. What will be the $pH$ of the solution?

Answer 1

Answer: 7.88

$C{H}_{3}COONa$ is salt of weak acid and strong base; its degree of hydrolysis may be calculated using the formula -
$\alpha =\sqrt{\left(\frac{K_h}{C}\right)}=\sqrt{\left(\frac{K_w}{C{K}_a}\right)}...(i)$
Hence, from Eq. (i)
$\alpha =\sqrt{\frac{10^{-14}}{10^{-3}\times 1.75\times 10^{-5}}}=7.55\times 10^{-4}$
$K_h$ (hydrolysis constant) $=\frac{K_w}{K_a}=\frac{10^{-14}}{1.75\times 10^{-5}}$
$=5.7\times 10^{-10}$
$pH$ after salt hydrolysis may be calculated as -
$pH=\frac{1}{2}\left[p{K}_w+p{K}_a+\log C\right]\dots$ (ii)
$p{K}_w=-\log K_w=-\log 10^{-14}=14$
$p{K}_a=-\log K_a=-\log \left(1.75\times 10^{-5}\right)$
$=4.7569$
$\log C=\log 10^{-3}=-3$
Substituting the values in Eq. (ii), we get
$pH=\frac{1}{2}[14+4.7569-31]$
or $pH=7.88$