Q. What is the pH of $10^{- 2}\ M\ H_{2}SO_{4}$?
A) 1.600
B) 2
C) $10^{- 2}$
D) 1.6990
Explanation
(d)
$10^{- 2}\ M\ H_{2}SO_{4} = 2 \times 10^{- 2}\ {NH}_{2}{SO}_{4}$
$\left\lbrack H^{+} \right\rbrack = 2 \times 10^{- 2}N$
$pH = - \log{\left\lbrack H^{+} \right\rbrack = - \log\left\lbrack 2 \times 10^{- 2} \right\rbrack}$
$= 2 - \log 2$
$= 2 - 0.3010$
$= 1.6990$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.