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Q. 100 mL of 0.015 M HCl solution is mixed "with 100 mL of 0.005 M HCl. What is the pH of the resultant solution?

EAMCETEAMCET 2004

Solution:

For $ \text{HCl} $ molarity is equal to normality. So, the normality of the resulting mixture is calculated as $ \text{N}\,\text{= }\frac{{{N}_{1}}{{V}_{1}}+N{{}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}} $ $ =\frac{0.015\times 100+0.005\times 100}{100+100} $ $ =\frac{1.5+0.5}{200}=\frac{2}{200}=\frac{1}{100}={{10}^{-2}} $ Normality of resulting mixture $ ={{10}^{-2}}N $ Resulting solution is acidic in nature. So, this resulting normality represent the $ [{{H}^{+}}] $ concentration. Then, $ [{{H}^{+}}]={{10}^{-2}} $ $ pH=-\log [{{H}^{+}}] $ $ =\log \frac{1}{[{{H}^{+}}]}=\log \frac{1}{{{10}^{-2}}} $ $ =\log {{10}^{2}} $ $ =2\log 10 $ $ =2 $