20 mL of 0.1 M H_2SO_4 is added to 30 mL | vedclass

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(D) Step $1$: Calculate millimoles $(m.mol)$ of reactants.$m.mol$ of $H_2SO_4 = 20 \ mL 	imes 0.1 \ M = 2 \ m.mol$.$m.mol$ of $NH_4OH = 30 \ mL 	ime

Vedclass · Accepted Answer

$9.4$

Vedclass · Answer

(D) Step $1$: Calculate millimoles $(m.mol)$ of reactants.$m.mol$ of $H_2SO_4 = 20 \ mL 	imes 0.1 \ M = 2 \ m.mol$.$m.mol$ of $NH_4OH = 30 \ mL 	imes 0.2 \ M = 6 \ m.mol$.Step $2$: Write the balanced chemical equation.$H_2SO_4 + 2NH_4OH 	o (NH_4)_2SO_4 + 2H_2O$.Step $3$: Determine the composition of the final mixture.Initial: $H_2SO_4 = 2 \ m.mol$,$NH_4OH = 6 \ m.mol$.After reaction: $H_2SO_4$ is the limiting reagent. $NH_4OH$ remaining $= 6 - (2 	imes 2) = 2 \ m.mol$. $(NH_4)_2SO_4$ formed $= 2 \ m.mol$.Step $4$: Calculate $pOH$ using the Henderson-Hasselbalch equation for a basic buffer.$pOH = pK_b + \log \frac{[Salt]}{[Base]} = 4.7 + \log \frac{2}{2} = 4.7 + 0 = 4.7$.Step $5$: Calculate $pH$.$pH = 14 - pOH = 14 - 4.7 = 9.3 \approx 9.4$.

$20 \ mL$ of $0.1 \ M \ H_2SO_4$ is added to $30 \ mL$ of $0.2 \ M \ NH_4OH$ solution. The $pH$ of the resultant mixture is [$pK_b$ of $NH_4OH = 4.7$].

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