Question

Calculate the pH of a buffer prepared by mixing $300$ cc of $0.3$ M $N{H}_3$ and $500$ cc of $0.5$ M $N{H}_{4}Cl$. $K_b$ for $N{H}_3=1.8\times {10}^{-5}$

• A. $8.1187$
• B. $9.8117$
• C. $8.8117$
• D. None of these

Answer 1

Answer: (C)

$8.8117$

Given$,$ $k_b$ for $N{H}_3=1.8\times {10}^{-5}$

Number of moles of $N{H}_3=M\times V$

$=M\times L$

$=0.3M\times 0.3L=0.09Moles$

Number of moles of $N{H}_{4}cl=M\times V=0.5M\times 0.5L=0.25moles$

$k_b$ for $N{H}_3=1.8\times {10}^{-5}$

$p{k}_b=-\log k_b=4.75$

$pOH$ $=p{K}_b+\log \frac{\left[salt\right]}{\left[base\right]}$

$pOH=4.75+\log \frac{0.25}{0.09}$

$=4.75-0.493$

$=4.307$

$pH=14-pOH$

$=14-4.307$

$=8.8117$

Hence$,$ Option $\left(C\right)$ is correct$.$