Question

what will be amount of (NH4)2SO4 (in g) which must be added to 500 mL of 0.2M NH4OH to yield a solution of pH 9.35?
Given,
pKa of NH4+ = 9.26
pKb of NH3 = 14-pKa(NH4+)​

Answer 1

$\sf pK_a$ of $\sf NH_4^+$ = 9.26

Hence,

$\sf pK_b$ of $\sf NH_4OH$ = 14 - 9.26 = 4.74

$\longrightarrow\sf\:pOH=14-9.35=4.65$

Each 1 mol of $\sf (NH_4)_2SO_4$ furnishes 2 moles of $\sf NH_4^+$ in solution.

Let $\sf [(NH_4)_2SO_4]$ = x mol L‾¹

$\sf[NH_4^+]$ = 2x mol L‾¹

$\sf[NH_4OH]$ = 0.2 mol L‾¹

Using Henderson - Hasselbalch equation,

$\sf:\implies\:pOH=pK_b+log\dfrac{[NH_4^+]}{[NH_4OH]}$

$\sf:\implies\:4.65=4.74+log\left(\dfrac{2x}{0.2}\right)$

This gives x = 0.081 mol L‾¹

Moles present in 500mL solution will be 0.081/2 = 0.0405

∴ Amount = 0.0405 × 132 = 5.35 g