Question

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 1

Answer: (C)

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, $[H^{+}]={10}^{-2}$ $pH=-\log [H^{+}]$ $=\log \frac{1}{[H^{+}]}=\log \frac{1}{{10}^{-2}}$ $=\log {10}^2$ $=2\log 10$ $=2$