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Chemistry Question on Acids and Bases

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

3.5

Explanation

Solution

∵ \because \, pH of HCl solution= 3.0 ∴    [H+] in HCl solution=1×10−3\therefore \, \, \, \, [H^+] \, in \, HCl \, solution = 1 \times 10^{-3} ∵ \because \, pH of NaOH solution= 10.0 ∴ [H+]i NaOH solution =1×10−141×10−10=10−4\therefore \, [H^+] i \, NaOH \, solution \, =\frac{1 \times 10^{-14}}{1 \times 10^{-10}} =10^{-4} Milliequivalents of HCl =N1V1=2.0×1×10−3N_1V_1 =2.0 \times 1 \times 10^{-3} \hspace30mm =2.0 \times 10^{-3} Milliequivalents of NaOH = 3.0×1×10−4=3.0×10−43.0 \times 1 \times 10^{-4} =3.0 \times 10^{-4} Since, milliequivalents of HCI are in excess, the milliequivalents of[H+H^+] in mixture        =(2.0×10−3)−(3.0×10−4) \, \, \, \, \, \, \, =(2.0 \times 10^{-3}) -(3.0 \times 10^{-4})          =1.7×10−3 \, \, \, \, \, \, \, \, \, =1.7 \times 10^{-3} Concentration of [H+^+] in mixture        =1.7×10−33+2=3.4×10−4 \, \, \, \, \, \, \, =\frac{1.7 \times 10^{-3}}{3+2}=3.4 \times 10^{-4} pH of mixture = −log[H+]-log[H^+]            =−log(3.4×10−4)=3.5 \, \, \, \, \, \, \, \, \, \, \, =-log (3.4 \times 10^{-4})=3.5