Q. Calculate the normality of 250 mL aqueous solution of $H_{2}{SO}_{4}$ having pH = 0.00.
A) 0.25 N
B) 0.50 N
C) 1 N
D) 2 N
Explanation
(b)
$pH\ = - log\lbrack H^{+}\rbrack$
$log\lbrack H^{+}\rbrack = - pH = 0.00$
[$H^{+}$]=antilog(0.00)
[$H^{+}$]= 1.0 M
$M\ H_{2}{SO}_{4}\ = \ 2{NH}_{2}{SO}_{4}$
$\therefore$ Normality of 250mL solution =$\frac{2 \times 250}{1000}$
=0.50 N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.