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

Answer:

  (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 \[N{{a}_{2}}S{{O}_{4}}=0.05\times 142=7.1g\] \[\text{Molarity}\,\text{of}\,\text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{=}\frac{\text{Number}\,\text{of}\,\text{moles}}{\text{Volume}\,\text{in}\,\text{litre}}\]                 \[=\frac{0.05}{2}=0.025mol\,{{L}^{-1}}\]