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1 M NaOH solution was slowly added to 1 L of 210 g impure H_(2)SO_(4) solution and the following plot was otained. Calculate the percentage purity of H_(2)SO_(4) sample. |
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Answer» Solution :From the plot, we observe that when 3 L of NaOH solution is consumed, `[H^(+)]` falls to zero. In other words, 3 Lof 1 M NaOH solution is used to nirtralize 1 L of the given `H_(2)SO_(4)` solution. But 3 L of 1 M NaOH contain 3 moles of NaOH and this will neutralize 1.5 mole of `H_(2)SO_(4)`. Thus, pure `H_(2)SO_(4)` present in 1 L of 210 g IMPURE `H_(2)SO_(4)` = 1.5 moles `=1.5xx98g=147g` `THEREFORE""%" purity"=(147)/(210)xx100=70%` 
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