Question

Calculate the extent of hydrolysis and the $pH$ of $0.02M$ ${CH}_{3}COO{NH}_4$.

$\left[K_b\right({NH}_3)=1.8\times {10}^{-5}$, $K_a\left({CH}_{3}COOH\right)=1.8\times {10}^{-5}$]

• A. $0.56$%, $pH=7$
• B. $0.76$%, $pH=12$
• C. $0.44$%, $pH=10$
• D. None of these

Answer 1

The correct option is A $0.56$%, $pH=7$

$C{H}_{3}COON{H}_4$ is a Salt of Weak Acid and Weak Base.
Hence,
$K_h$$=\frac{K_w}{K_a{K}_b}$$=\frac{{10}^{-14}}{1.8\times {10}^{-5}\times 1.8\times {10}^{-5}}$$\approx 3.1\times {10}^{-5}$
Also, we know that
$K_h\approx h^2$
$h=$$\sqrt{3.1\times {10}^{-5}}=5.56\times {10}^{-3}$

$\%h=$$h\times 100=0.56$%

Since for the given salt, $K_a=K_b$, hence,the solution of this salt in water is neutral and its $pH$=7.