Question

10 mL of 0.1 N $HCl$ is added to 990 mL solution of $NaCl$. The pH of the resulting solution is:

• A. zero
• B. 3
• C. 7
• D. 10

Answer 1

The correct option is B 3

For HCl, Molarity = Normality (since n-factor is 1)

Number of moles of $H^{+}=M\times V=0.1\times 10\times {10}^{-3}={10}^{-3}$

Total volume $=10+990ml=1litre$

$\left[H^{+}\right]=\frac{{10}^{-3}}{1}M={10}^{-3}M$

$pH=-log{10}^{-3}=3$