Question

Calculate the pH of a 0.033 M ammonia solution, if 0.33 M $N{H}_{4}Cl$ is introduced in this solution at the same temperature. $(k_{b}forN{H}_3=1.77\times {10}^{-5})$

• A. 4.75
• B. 10
• C. 9.25
• D. None of these

Answer 1

Answer: (C)

9.25

We know,

$pOH=p{K}_b+\log \left(\right[Salt\left]/\right[Base\left]\right)$

where,

$K_b$ for $N{H}_3=1.77\times {10}^{-5}$

$p{K}_b=-\log K_b=-\log (1.77\times {10}^{-5})$

$\left[Salt\right]=\left[N{H}_{4}Cl\right]=0.033M$

$\left[Base\right]=\left[N{H}_3\right]=0.033M$

So,

$pOh=4.75+\log \frac{0.033}{0.033}$

$pOH=4.75$

$\therefore pH=14-4.75=9.25$