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Sulphuric acid reacts with sodium hydroxide as follows : H_(2)SO_(2)+2NaOH - Na_(2)SO_(4)+2H_(2)O When 1 L of 0.1 M sulphuric acid solution is allowed to react will 1L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is |
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Answer» `"0.1 mol L"^(-1)` `"1 L of 0.1 M NaOH contains = 0.1 mole of NaOH"` According to the given equation, 1 mole of `H_(2)SO_(4)` reacts with 2 moles of NaOH. Hence 0.1 mole of NaOH will REACT with 0.05 mole of `H_(2)SO_(4)` (and 0.05 mole of `H_(2)SO_(4)` will be left unreacted), i.e., NaOH is the limiting reactant.. 2 moles of NaOH produce 1 mole of `Na_(2)SO_(4)`. Hnece 0.1 mole of NaOH will produce 0.05 mole of `Na_(2)SO_(4)` `=0.05xx(46+32+64)g=0.05xx142g=7.10g` Volume of solution after MIXING = 2 L `H_(2)SO_(4)` left unreacted in the solution = 0.05 mole `therefore"Molarity of the solution "=(0.05)/(2)=0.025" mol L"^(-1)` |
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