Which of the following will decrease the $p H$ of $50 m l$ solution of $0.01 M H C l$ ?

Addition of

M g

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Addition of

10 m l, 0.5 M H C l

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Addition of

100 m l, 0.001 M H C l

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Addition of

50 m l, 0.002 M H C l

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Solution

The correct option is B Addition of $10 m l, 0.5 M H C l$
Initial $p H = - l o g [H^{+}]$
$= - l o g [10^{- 2}]$
$p H = 2$
$N = 0.01 N$
Now, $N^{1} = \frac{N_{1} V_{1} + N_{2} V_{2}}{V_{1} + V_{2}}$
$= \frac{(50) (0.01) + (10) (0.5)}{60} = \frac{11}{120} {p H}^{'} = 1.037$
$p H$ decreases by addition of $10 m l, 0.5 M; H C l$

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