Question

Calculate the extent of hydrolysis and the pH of 0.02 M CH3COONH4. Kb(NH3)=1.8*10^-5,Ka(CH3COOH)=1.8*10^-5.

Answer 1

Answer: The extent of hydrolysis is $0.27\times 10^{-9}$ and the pH of the solution is 7.

Explanation:

$CH_3COONH_4+H_2O\rightarrow CH_3COOH+NH_4OH$

$h=\sqrt{\frac{K_w}{K_a+K_b}}$

h = degree f hydrolysis for salt of weak acid and weak base

$K_a=1.8\times 10^{-5}$

$K_b=1.8\times 10^{-5}$

$K_w=10^{-14}$

$h=\sqrt{\frac{10^{-14}}{1.8\times 10^{-5}+1.8\times 10^{-5}}}=0.27\times 10^{-9}$

2) $pH=\frac{1}{2}\times (pK_w+pK_a - pK_b)$

$pK_w=-log(K_w)=-log[10^{-14}]=14$

$pK_a=-log(K_a)=-log[1.8\times 10^{-5}]=4.75$

$pK_w=-log(K_b)=-log[1.8\times 10^{-5}]=4.75$

$pH=\frac{1}{2}\times (14+4.75-4.75)=7$