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# $pH$ of the solution containing 50.0 mL of 0.3 M HCl and 50.0 mL of 0.4 M $NH_3$ is

[Given $pK_a (NH^+_4)$ = 9.26]

(a) 4.74

(b) 9.26

(c) 8.78

(d) 4.63

Toolbox:
• $pOH = pK_b + log \frac{[Salt]}{[Base]}$

$pK_a(NH^+_4) = 9.26$

$\therefore pK_b (NH_3) = pK_w - pK_a(NH^+_4)$ = 14 - 9.26 = 4.74

On adding 15 m mol (50 mL of 0.3 M) HCl to 20 m mol (50 mL of 0.4 M) $NH_3$ the resulting solution contains 15 m mol (in 100 mL) of $NH_4Cl$ and 5 m mol (in 100 mL) of $NH_3$.
$\therefore [Salt] = 15 \times 10^{-2} M , [Base] = 5 \times 10^{-2} M$

$pOH = pK_b + log \frac{[Salt]}{[Base]} = 4.74 + log \frac{15 \times 10^{-2}}{5 \times 10^{-2}} = 4.74 + log 3 = 4,74 + 0.48 = 5.22$

pH = 14 - pOH = 14 - 5.22 = 8.78