Question

200 mL of a $\frac{N}{10}$ $H_{2}S{O}_4$ solution is mixed with 300 mL of a $\frac{N}{100}$ NaOH solution. Calculate the normality of the excess $H^{+}/O{H}^{-}$ ions in the resulting solution.

• A. 0.024 N
• B. 0.015 N
• C. 0.050 N
• D. 0.034 N

Answer 1

The correct option is D 0.034 N

Number of gram equivalents of $H_{2}S{O}_4=$ Normality $\times$ Volume in $L=0.1\times 0.2=0.02$
Number of gram equivalents of NaOH = Normality $\times$ Volume in L
= 0.01 $\times$ 0.3 = 0.003
0.003 gram equivalents of NaOH will neutralize 0.003 gram equivalents of $H_{2}S{O}_4.$
Hence, the number of gram equivalents of $H_{2}S{O}_4$ in 500 mL of the solution after neutralisation = (0.02-0.003) = 0.017
Hence, the number of gram equivalents of $H^{+}$= 0.017
Normality of $H^{+}$= $\frac{\text{Number of gram equivalents}}{\text{Volume of Solution in L}}$
Normality of $H^{+}$= $\frac{0.017}{0.5L}=0.034$ N