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Law of Mass Action

50 mL of ...

Question

$50 m L$ of $0.1 M H C l$ and $50 m L$ of $0.2 M N a O H$ are mixed. The pH of the resulting solution is

1.30

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4.2

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12.70

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11.70

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Solution

The correct option is C $12.70$
$50 m L$ of $0.1 M H C l = 50 \times 0.1$ millimoles $= 5$ millimoles.
$50 m L$ of $0.2 N a O H = 50 \times 20$ millimoles $= 10$ millimoles.
Millimoles of $N a O H$ left after neutralization $= 5$ .
Volume of solution $= 100 m L$ .
$∴$ Molar conc. of $N a O H = \frac{5}{100} = 5 \times 10^{- 2} M$
$p O H = - l o g (5 \times 10^{- 2})$
$= l o g \frac{1}{5 \times 10^{- 2}} = l o g 20 = 1.301$
$p H = 14 - 1.301 = 12.699 = 12.70$

Hence, Option "C" is the correct answer.

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