• # question_answer 61)   Sulphuric acid reacts with sodium hydroxide as follows : ${{H}_{2}}S{{O}_{4}}+2NaOH\to N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O$ When 1 L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is: (a) $0.1mol{{L}^{-1}}$                  (b) 7.10 g (c) $0.025mol{{L}^{-1}}$             (d) 3.55 g

(b, c) ${{H}_{2}}S{{O}_{4}}+2NaOH\to N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O$ $\frac{{{M}_{1}}{{V}_{1}}}{{{n}_{1}}}({{H}_{2}}S{{O}_{4}})=\frac{{{M}_{2}}{{V}_{2}}}{{{n}_{2}}}(NaOH)$ or      $\frac{0.1\times {{V}_{1}}}{1}=\frac{0.1\times 1}{2}$ or            ${{V}_{1}}=0.5L=500mL$ Thus, number of moles of ${{H}_{2}}S{{O}_{4}}$ used = $\frac{MV}{1000}$                                                 $=\frac{0.1\times 500}{1000}=0.05$ Number of moles of $N{{a}_{2}}S{{O}_{4}}$ formed will also be equal to 0.05. Mass of $N{{a}_{2}}S{{O}_{4}}=0.05\times 142=7.1g$ $\text{Molarity}\,\text{of}\,\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{=}\frac{\text{Number}\,\text{of}\,\text{moles}}{\text{Volume}\,\text{in}\,\text{litre}}$                 $=\frac{0.05}{2}=0.025mol\,{{L}^{-1}}$