Q. What will be the pH value of 0.05M $Ba(OH)_{2}$ solution?
A) $12$
B) $13$
C) $1$
D) 12.96
Explanation
(b)
0.05 M $Ba(OH)_{2}$ solution
$\cong 2 \times 0.05\ N\ Ba(OH)_{2}$
$\cong 0.10\ N\ Ba(OH)_{2}$
$\therefore$ $pOH = - \log{\lbrack OH^{-}\rbrack}$
$= - \log{(0.10) = 1}$
$pH = 14 - pOH$
$= 14 - 1 = 13$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.