Question

pH of $0.01M(N{H}_4{)}_{2}S{O}_4$ and $0.02MN{H}_{4}OH$ buffer $(p{K}_{a}ofN{H}_4^{+}=9.26)$ is

• A. 4.74 + log 2
• B. 4.74 - log 2
• C. 9.26 +log 2
• D. 9.26 + log 1

Answer 1

Answer: (D)

$p{K}_a\left(N{H}_4^{+}\right)=9.26$
$p{K}_b\left(N{H}_3\right)=14-9.26=4.74$
$\left[N{H}_4^{+}\right]=2\times \left[{\left(N{H}_4\right)}_{2}S{O}_4\right]$
$=2\times 0.01=0.02M$
$pOH=p{K}_b+log\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}$
$pOH=4.74+log\frac{0.02}{0.02}=4.74-log1$
$pH=14-pOH=14-4.74+log1$
$=9.26+log1$