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Sulphuricacid reacts with sodium hydroxide as follows : H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) +2H_(2)O When 1 L of 0.1M sulphuric acid solution is allowed to react with 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)` `H_(2)SO_(4) +2NaOH rarr Na_(2)SO_(4) +2H_(2)O` 1 L of 0.1M `H_(2)SO_(4)` contains `=0.1` mole of `H_(2)SO_(4)` 1 L of 0.1 M NaOH contains `=0.1` mole of `H_(2)SO_(4)` ACCORDING to the reaction, 1 mole of `H_(2)SO_(4)` reacts with 2 MOLES of NaOH will react with 0.5 mole `H_(2)SO_(4)` and 0.05 mole of `H_(2)SO_(4)` will remain unreacted i.e., NaOH is the limiting reactant. Since, 2 moles of NaOH produces 1 mole of `Na_(2)SO_(4)`. Hence, 0.1 mole of NaOH will produce 0.05 mole of `Na_(2)SO_(4)`. Mass of `Na_(2)SO_(4) = "moles" xx "molar mass"` `= 0.5xx(46+32+64)g` `= 7.10g` Volume of solution after mixing `=2L` Since, only 0.05 mole of `H_(2)SO_(4)` is left behind as NaOH completely used in this reaction. Therefore, molarity of the given solution is calculated from moles of `H_(2)SO_(4)`. `H_(2)SO_(4)` left unreacted in the solution `= 0.05` mole Molarity of the solution `= (0.5)/(2) = 0.025 "mol" L^(-1)` |
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