Question

What will be the amount of $(NH_4)_2SO_4$ (in g) which must be added to $500\,mL$ of $0.2\,M \,NH_4OH$ to yield a solution of $pH \,9.35$? [Given, $pK_a$ of $NH_4^+ = 9.26$, $pK_b\,NH_4OH = 14 - pK_a(NH_4^+)]$

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

Answer 1

Answer: (A)

$pK_a$ of $NH^+_4 = 9.26$
Hence, $pK_b$ of $NH_4OH = 14 - 9.26 = 4.74$
$pOH = 14 - 9.35 = 4.65$
Each $1 \,mol$ of $(NH_4)_2SO_4$ furnishes $2 \,moles$ of $NH^+_4$ in solution.
Let $[(NH_4)_2SO_4] = x \,mol \,L^{-1}$
$[NH^+_4] = 2\, x\,mol\,L^{-1}$
$[NH_4OH] = 0.2 \,mol\, L^{-1}$
using Henderson-Hasselbalch equation,
$pOH=pK_{b}+log \frac{\left[NH^{+}_{4}\right]}{\left[NH_{4}OH\right]}$
$4.65=4.74+log\left(\frac{2x}{0.2}\right)$
This gives $x = 0.081 \,mol\, L^{-1}$
$= 0.0405 \,mol$ in $500 \,mL$ (as required)
$= 0.0405 \times 132 = 5.35 g$