Question

15 mL of a $\frac{N}{10}$ NaOH solution completely neutralizes 12 mL of a $H_{2}S{O}_4$ solution. The normality of the $H_{2}S{O}_4$ solution will be:

• A. $\frac{N}{5}$
• B. $\frac{N}{10}$
• C. $\frac{N}{8}$
• D. N

Answer 1

The correct option is C $\frac{N}{8}$

For complete neutralisation, the number of equivalents of acid should be equal to the number of equivalents of the base.
$\Rightarrow N_1{V}_1=N_2{V}_2$
Where,
$N_1\Rightarrow$Normality of the NaOH solution = $\frac{1}{10}$ N
$V_1\Rightarrow$ Volume of NaOH solution = 15 mL
$N_2\Rightarrow$ Normality of the $H_{2}S{O}_4$ solution
$V_2\Rightarrow$ Volume of $H_{2}S{O}_4$ solution = 12 mL
$\frac{1}{10}\times 15=N_2\times 12$
$N_2=\frac{1}{8}$ N