A) It must be increased 10 times
B) It must be increased by 1 mole/litre
C) It must be decreased to \[\frac{1}{10}\]th of its original value
D) It must be decreased by 1 mole/litre
Correct Answer: C
Solution :
We know that, \[pH=\log \frac{1}{[{{H}^{+}}]}\] Hence, \[{{(pH)}_{1}}=-\log \,{{[{{H}^{+}}]}_{1}}\] ?(i) \[{{(pH)}_{2}}=-\log \,{{[{{H}^{+}}]}_{2}}\] ?(ii) Subtract eq. (i) from equation (ii): \[{{(pH)}_{2}}--{{(pH)}_{1}}=-\log \,{{[{{H}^{+}}]}_{2}}+\log \,{{[{{H}^{+}}]}_{1}}\] or \[(pH)+1--{{(pH)}_{1}}=\log \frac{{{[{{H}^{+}}]}_{1}}}{{{[{{H}^{+}}]}_{2}}}\] or \[1=\log {{[{{H}^{+}}]}_{1}}/{{[{{H}^{+}}]}_{2}}\] or \[\frac{{{[{{H}^{+}}]}_{1}}}{{{[{{H}^{+}}]}_{2}}}=10\] \[(\because \,1=\log 10)\] Hence, \[{{[{{H}^{+}}]}_{2}}=\frac{1}{10}{{[{{H}^{+}}]}_{1}}\] i.e. \[[{{H}^{+}}]\] must be decreased \[\frac{1}{10}th\]of its original volume.
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