Question

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A mixture of H2SO4 and H2C2O4 and some inert impurity weighing 3.185 g was dissolved in water and made up to 1 L . 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment 100mL of the same solution in hot condition required 4mL of 0.02 M KMnO4 solution for complete reaction. The mass % of H2SO4 in the mixture was : ?
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Answer 1

Answer:

We have to find the moles of KMNO4.

Now to find the moles we know 0.004 L of 0.02 M KMnO4 hence  0.00008 mol KMnO4  will form.

So, 0.00008 mol KMnO4 * 5 mol H2C2O4 / 2 mol KMnO4 = 0.0002 mol of oxalic acid will form per 100 ml of the  mixture.  

Now, the moles of NaOH.  

So, 0.003 L of 0.1 N = 0.0003 mol of the  NaOH  will form.

Hence, 0.0003 mol * 1 mol of the acid / 2 mol NaOH = 0.00015 mol of acid will forms per 10 ml of the  mixture.  

Which means 0.0015 mol acids per 100 ml of the mixture.  

So, total 0.0015 mol acids - 0.0002 mol of oxalic acid = 0.0013 mol diprotic acid per 100 ml of the mixture or 0.013 mol sulfuric acid per litre of mix ture.

So, 0.013 mol H2SO4 *  98.08 g/mol will be 1.275 grams of H2SO4.

So, % mass of H2SO4 will be 1.275 grams of H2SO4 / 3.185 g of  mixture = 40.03 % H2SO4.