Question

What will be the amount of $(N{H}_4{)}_{2}S{O}_4$ (in g) which must be added to 500 mL of 0.2 M $N{H}_{4}OH$ to yield a solution of pH 9.35? [Given, $p{K}_a$ of $N{H}_4^{+}$ = 9.26, $p{K}_b\left(N{H}_3\right)$ = 14 - $p{K}_a\left(N{H}_4^{+}\right)$]

• A. 5.35
• B. 6.47
• C. 10.03
• D. 7.34

Answer 1

The correct option is A 5.35

$p{K}_a$ of $N{H}_4^{+}=9.26$

Hence, $p{K}_b$ of $N{H}_{4}OH=14-9.26=4.74$

pOH = $14-9.35=4.65$

Each $1$ ml of $(N{H}_4{)}_{2}S{O}_4$ furnishes $2$ moles of $N{H}_4^{+}$ in solution.

Let $\left[\right(N{H}_4{)}_{2}S{O}_4]=x$ $mol{L}^{-1}$

$\left[N{H}_4^{+}\right]=2x$ $mol{L}^{-1}$

$\left[N{H}_{4}OH\right]=0.2$ $mol{L}^{-1}$

using Henderson - Hasselbalch equation,

$pOH=p{K}_b+log\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}$

$4.65=4.74+log\left(\frac{2x}{0.2}\right)$

This gives $x=0.081$ $mol{L}^{-1}$

=$0.0405$ mol in 500mL (as required)

weight=$0.0405\times 132=5.35g$