Equal volumes of $0.02 M H C l$ and $0.04 M N a O H$ solutions are mixed. The pH of the mixture is:

10

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2

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7

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12

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Solution

The correct option is B $12$
It is given that equal volumes of $0.02 M H C l$ and $0.04 M N a O H$ solutions are mixed.

Let us assume that $x$ L each of $H C l$ and $N a O H$ solutions are mixed.
The formula for the hydroxide ion concentration is $[O H]^{-} = \frac{M_{b} V_{b} - M_{a} V_{a}}{V_{a} + V_{b}}$
On substituting values in the above expression, we get
$[O H]^{-} = \frac{0.04 \times x - 0.02 \times x}{x + x} = \frac{0.02 x}{2 x} = 0.01 M$
$p O H = - l o g [O H]^{-} = - l o g (0.01) = 2$
$p H = 14 - p O H = 14 - 2 = 12$

Hence, the pH of the mixture is $12$ .

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