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

$CH _{3} COONa$ is salt of weak acid and strong base; its degree of hydrolysis may be calculated using the formula -
$\alpha=\sqrt{\left(\frac{ K _{ h }}{ C }\right)}=\sqrt{\left(\frac{ K _{ w }}{ CK _{ a }}\right)}\,\,\,...(i)$
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 -
$pH =\frac{1}{2}\left[ pK _{ w }+ pK _{ a }+\log C \right] \ldots$ (ii)
$pK _{ w }=-\log K _{ w }=-\log 10^{-14}=14$
$pK _{ a }=-\log K _{ a }=-\log \left(1.75 \times 10^{-5}\right)$
$=4.7569$
$\log C=\log 10^{-3}=-3$
Substituting the values in Eq. (ii), we get
$pH =\frac{1}{2}[14+4.7569-31] $
or $ pH =7.88$