Question

1 cc of 0.1N $HCl$ is added to 999 cc solution of $NaCl$. The $pH$ of the resulting solution will be:

• A. 7
• B. 3
• C. 4
• D. 1

Answer 1

Answer: (C)

4

Given

$V_1=1c{m}^3=0.001L{V}_2=999c{m}^3=0.999L{N}_1=0.1N$

Total volume of solution $=V_1+V_2=1L$

$NaCl$ is a salt and does not affect $pH$ of a solution.

Now, normality of $HCl$ is resulting solution,

$N_1{V}_1=N_2{V}_2$

$\therefore 0.1\times 0.001=N_2\times 0.999$

$\therefore N_2=\frac{0.0001}{1}=0.0001N=1\times {10}^{-4}$

$\therefore$ $pH=-\log \left[H^{+}\right]=-\log [1\times {10}^{-4}]$

$\therefore pH=4$

Hence, option $C$ is correct.