Q. Given pH of a solution $A$ is 3 and it is mixed with another solution $B$ having pH 2. After mixing are resultant pH of the solution will be
A) 3.2
B) 1.9
C) 3.4
D) 3.5
Explanation
(b)
pH of the solution $A = 3$
$\left\lbrack H^{+} \right\rbrack_{A} = 10^{- 3}$ M
pH of the solution $B = 2$
$\left\lbrack H^{+} \right\rbrack_{B} = 10^{- 2}$ M
$\left\lbrack H^{+} \right\rbrack = 10^{- 3} + 10^{- 2} = 11 \times 10^{- 3}$
$pH = - \log{\left( 11 \times 10^{- 3} \right) = 3 - \log 11}$
$= 3 - 1.04 = 1.9$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.