Question

If $p{K}_a\left(N{H}_4^{+}\right)=9.2$, then the amount of $(N{H}_4{)}_{2}S{O}_4$ to be added in 500 mL of 0.1 M $N{H}_{4}OH$ for the preparation of a buffer of pH=8.5 would be:
(Given: log5=0.7)

• A. 0.175 mol
• B. 0.125 mol
• C. 0.150 mol
• D. 0.20 mol
  

Answer 1

The correct option is B 0.125 mol

This will make a basic buffer.
So, according to the Henderson- Hasselbalch equation,
$pOH=p{K}_b+log\frac{\left[salt\right]}{\left[base\right]}$
So, $5.5=4.8+log\frac{n_{salt}}{0.05}$
$0.7=log\frac{n_{salt}}{0.05}$
$\Rightarrow log\frac{\left(n_{salt}\right)}{0.05}=log5$
$\Rightarrow n_{salt}=0.25mol$
Since $(N{H}_4{)}_{2}S{O}_4$ includes 2 $N{H}_4^{+}$ ions, the amount of $(N{H}_4{)}_{2}S{O}_4$ required will be 0.125 mol.