Question

Calculate hydrolysis constant for 0.01 (N) solution of $C{H}_{3}C{O}_{2}Na$ ($K_a$ for $C{H}_{3}C{O}_{2}H=1.8\times {10}^{-5}$).

Answer 1

For $C{H}_{3}COONa$ the equivalent wt. = molecular wt. so the concentration can taken as 0.01 M

(For salts equivalent mass = formula mass/total charge of cation or anion.)

Reaction of $C{H}_{3}COONa$ in water can be written as :

$C{H}_{3}COONa+H_{2}O\to C{H}_{3}COOH+N{a}^{+}+O{H}^{-}$

So we need to calculate the conc. of $O{H}^{-}$ to get pH of salt solution.

For calculating pH , at first we will find out the value of Kb using Ka

$K_w=K_a{K}_b$

$1.00\times {10}^{-14}=(1.80\times {10}^{-5})\left(K_b\right)$

$K_b=5.56\times {10}^{-10}$

Use the Kb expression to calculate the $\left[O{H}^{¯}\right]:$

$K_b=\left[N{a}^{+}\right]\left[O{H}^{-}\right]/C{H}_{3}COONa$ ........(1)

Let the change in concentration after dissociation is x, so the conc. of $N{a}^{+},O{H}^{-}$ would be x. Putting the valuesin eq (1) we get :

$5.56\times {10}^{-10}=\left[\right(x\left)\right(x\left)\right]/(0.01-x)$

neglect the minus x

$x=2.3\times {10}^{-6}M=\left[O{H}^{-}\right]=\left[N{a}^{+}\right]$ Calculate pOH, then pH:

$pOH=-log[2.3\times {10}^{-6}M]=5.6$

$pH=14-pOH=14-5.6=8.4$