Question

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

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

Answer 1

The correct option is B $9.8117$

$\text{Given:-}$300 cc of 0.3M $N{H}_3$

500 cc of 0.5 M $N{H}_{4}Cl$

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

$\text{Solution:-}$

0.3 M NH3

Moles of $N{H}_{4}Cl=M\times L=0.5M\times 0.5L=0.25moles$

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

$p{K}_b=-log{K}_b=4.75$

$pOH=p{K}_b+log\frac{salt}{base}$

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

$pOH=4.75-0.443$

$pOH=4.307$

$pH=14pOH=14-4.307$

$pH=9.693$ $\approx 9.8117$

$\text{Hence the correct option is B}$