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  4. How much the hydrogen ion concentration in a solution be changed for raising the pH by 1 unit?

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Q. How much the hydrogen ion concentration in a solution be changed for raising the pH by 1 unit?

JIPMERJIPMER 1998

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.