Question

The pH of a solution prepared by mixing 2.0 mL of HCl solution of pH 3.0 and 3.0 mL of NaOH of pH 10.0 is

• A. 2.5
• B. 3.5
• C. 5.5
• D. 6.5

Answer 1

The Correct Option is B

$\because \, $ pH of HCl solution= 3.0 $\therefore \, \, \, \, [H^+] \, in \, HCl \, solution = 1 \times 10^{-3}$ $\because \, $ pH of NaOH solution= 10.0 $\therefore \, [H^+] i \, NaOH \, solution \, =\frac{1 \times 10^{-14}}{1 \times 10^{-10}} =10^{-4}$ Milliequivalents of HCl =$N_1V_1 =2.0 \times 1 \times 10^{-3}$ $\hspace30mm =2.0 \times 10^{-3}$ Milliequivalents of NaOH = $3.0 \times 1 \times 10^{-4} =3.0 \times 10^{-4}$ Since, milliequivalents of HCI are in excess, the milliequivalents of[$H^+$] in mixture $ \, \, \, \, \, \, \, =(2.0 \times 10^{-3}) -(3.0 \times 10^{-4})$ $ \, \, \, \, \, \, \, \, \, =1.7 \times 10^{-3}$ Concentration of [H$^+$] in mixture $ \, \, \, \, \, \, \, =\frac{1.7 \times 10^{-3}}{3+2}=3.4 \times 10^{-4}$ pH of mixture = $-log[H^+]$ $ \, \, \, \, \, \, \, \, \, \, \, =-log (3.4 \times 10^{-4})=3.5$