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Chemistry Question on Acids and Bases

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?

A

2.5

B

1.5

C

2

D

1

Answer

2

Explanation

Solution

For HCl\text{HCl} molarity is equal to normality. So, the normality of the resulting mixture is calculated as N = N1V1+N2V2V1+V2\text{N}\,\text{= }\frac{{{N}_{1}}{{V}_{1}}+N{{}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}} =0.015×100+0.005×100100+100=\frac{0.015\times 100+0.005\times 100}{100+100} =1.5+0.5200=2200=1100=10−2=\frac{1.5+0.5}{200}=\frac{2}{200}=\frac{1}{100}={{10}^{-2}} Normality of resulting mixture =10−2N={{10}^{-2}}N Resulting solution is acidic in nature. So, this resulting normality represent the [H+][{{H}^{+}}] concentration. Then, [H+]=10−2[{{H}^{+}}]={{10}^{-2}} pH=−log⁡[H+]pH=-\log [{{H}^{+}}] =log⁡1[H+]=log⁡110−2=\log \frac{1}{[{{H}^{+}}]}=\log \frac{1}{{{10}^{-2}}} =log⁡102=\log {{10}^{2}} =2log⁡10=2\log 10 =2=2