The pH of a solution obtained by mixing $5$ mL of $0.1$ M $NH_4OH$ solution with $250$ mL of $0.1$ M $NH_4Cl$ solution is _____ $\times 10^{-2}$ . (Nearest integer)
Given: $pK_b(NH_4OH) = 4.74$
$\log 2 = 0.30$
$\log 3 = 0.48$
$\log 5 = 0.70$

Question

The pH of a solution obtained by mixing 5 mL of 0.1 M NH_4OH solution with 250 mL of 0.1 M NH_4Cl solution is _____ × 10^ -2 . (Nearest integer) Given: pK_b(NH_4OH) = 4.74 2 = 0.30 3 = 0.48 5 = 0.70

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Correct answer: 0

The given mixture forms a basic buffer solution.

Millimoles of base

NH_4OH = 5 \times 0.1 = 0.5

mmol

Millimoles of salt

NH_4Cl = 250 \times 0.1 = 25

mmol

Using the Henderson-Hasselbalch equation for a basic buffer:

pOH = pK_b + \log \left( \dfrac{[\text{Salt}]}{[\text{Base}]} \right)

pOH = 4.74 + \log \left( \dfrac{25}{0.5} \right)

pOH = 4.74 + \log (50)

We know that

\log (50) = \log (5 \times 10) = \log 5 + \log 10 = 0.70 + 1 = 1.70

pOH = 4.74 + 1.70 = 6.44

The pH of the solution is given by:

pH = 14 - pOH

pH = 14 - 6.44 = 7.56

pH = 756 \times 10^{-2}

Answer:

756

The pH of a solution obtained by mixing 5 mL of 0.1 M NH_4OH...

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