Question

Find the purity percentage of \[{H_2}S{O_4}\] with density \[1.8\] , if \[5\] ml of \[{H_2}S{O_4}\] is neutralized completely by \[84.6\] ml of \[2N\; NaOH\] solution.

Answer 1

Hint: The purity percentage can be obtained by the mass of the solution and mass of sulfuric acid.
The mass of the sulfuric acid can be calculated from the given values. The mass of the solution can be determined from density and volume of solution i.e.., \[1\] litre.

Formula Used:
\[{N_1}{V_1} = {N_2}{V_2}\]
\[{N_1}\] is Normality of sulfuric acid \[\left( {{H_2}S{O_4}} \right)\]
\[{V_1}\] is volume of sulfuric acid \[\left( {{H_2}S{O_4}} \right)\]
\[{N_2}\] is Normality of sodium hydroxide \[\left( {NaOH} \right)\]
\[{V_2}\] is volume of sodium hydroxide \[\left( {NaOH} \right)\]

Complete answer: Given that, the density of sulfuric acid \[\left( {{H_2}S{O_4}} \right)\]is \[1.8\]
\[{V_1}\] is volume of sulfuric acid \[\left( {{H_2}S{O_4}} \right)\]given as \[5\] ml
\[{N_2}\]is Normality of sodium hydroxide \[\left( {NaOH} \right)\] given as \[2N\]
\[{V_2}\] is volume of sodium hydroxide \[\left( {NaOH} \right)\] given as \[84.6\] ml
By substituting the values in the above formula:
\[{N_1} \times 5 = 2 \times 84.6\]
Thus, the Normality of sulfuric acid \[\left( {{H_2}S{O_4}} \right)\] will be
\[{N_1} = \left( {84.6 \times 2} \right) \div 5 = 33.84N\]
Thus, the \[M\] will be
 \[N \div \text{Valency Factor}\]
Substitute the value of valency factor in the above equation
\[33.84 \div 2 = 16.92M\]
It means \[16.92M\] moles of \[{H_2}S{O_4}\] is present in \[1000\] ml of the solution
But, mass of solution can be obtained from the density and volume of solution
Thus, \[mass = density \times volume\]
Substitute the values of density and volume of solution is \[1000\] ml
Thus, mass of solution will be \[1.8 \times 1000 = 1800gm\]
Mass of the sulfuric acid will be \[16.92 \times 98 = 1658.16gm\]
The percentage purity of sulfuric acid \[\left( {{H_2}S{O_4}} \right)\] will be
\[\dfrac{{1658.16}}{{1800}} \times 100 = 92.12\% \]
Thus, the percentage purity of sulfuric acid \[\left( {{H_2}S{O_4}} \right)\] is approximately \[92\% \].

Note:
The number of moles can be calculated from the molar mass and mass or amount of the substance.
The molar mass of sulfuric acid \[\left( {{H_2}S{O_4}} \right)\] is \[98gm{\left( {mol} \right)^{ - 1}}\]
Valency factor is given by the number of protons replaceable in the given acid.