Question

The specific gravity of a given \[{{\text{H}}_{2}}{\text{S}}{{\text{O}}_4}\] solution is \[1.76\] . A quantity of \[3.5\] ml of the acid is diluted to \[1.0\] L and 25 ml of this diluted acid required \[25.6\] ml of \[\dfrac{{\text{N}}}{{10}}\left( {{\text{f}} = 0.95} \right){\text{NaOH}}\] solution for complete neutralization. The percentage strength (by mass) of the original acid solution is:
A.\[61.6\% \]
B.\[77.38\% \]
C\[{\text{50\% }}\]
D.\[{\text{47}}{\text{.66\% }}\]

Answer 1

Hint: The reaction of an acid and a base to form salt and water is called neutralisation. First of all we shall calculate the milliequivalents of sulphuric acid and then its mass can be calculated. Then using the density mass of solution and hence percentage composition can be calculated.

Complete step by step answer:
First of all we will calculate milliequivalents of \[{\text{NaOH}}\] by multiplying the normality with volume given.
The number of mili equivalent of \[{\text{NaOH}}\] will be \[\dfrac{{\text{N}}}{{10}} \times 25.6 = 2.56\]
But the actual reacting milliequivalents will be calculated with the help of n factors given to us. We will multiply n factor with the mili equivalent \[2.56 \times 0.95 = 2.432{\text{ m eq}}\]
The neutralisation between sulphuric acid and sodium hydroxide occurs as:
\[{{\text{H}}_2}{\text{S}}{{\text{O}}_4} + 2{\text{NaOH}} \to {\text{N}}{{\text{a}}_{2}}{\text{S}}{{\text{O}}_4} + 2{{\text{H}}_2}{\text{O}}\]
For 2 moles of sodium hydroxide 1 mole of sulphuric acid is required.
Hence, for \[2.432{\text{ m eq}}\] of sodium hydroxide \[\dfrac{{2.432{\text{ m eq}}}}{2} = 1.216\] moles of sulphuric acid will be used.
Now the volume for neutralisation of acid is 25 mL and the number of moles we have calculated.
For 25 mL acid the moles are \[1.216\]
For 1 L or 1000 mL acid the moles of acid will be \[\dfrac{{1.216}}{{25}} \times 1000 = 48.64{\text{ moles}}\]
Now the volume of the acidic solution is \[3.5\] mL and the number of moles are \[48.64\] .
The molarity of the acid will be: \[\dfrac{{48.64}}{{3.5}} = 13.89\]
Mass of sulphuric acid can be calculated by multiplying it with molar mass.
Mass of sulphuric acid: \[13.89 \times 98 = 1361{\text{gm}}\]
Now the density of solution is given to us and the volume of solution is 1000 mL, so the mass of solution will be:
\[1000{\text{d}} = 1000 \times 1.76 = 1760{\text{ g}}\]
The percentage by mass is defined as the mass of solute in gram present in 100 g of solution.
\[\% \;{{\text{H}}_{2}}{\text{S}}{{\text{O}}_4} = \dfrac{{1361}}{{1760}} \times 100 = 77.32\% \]

Hence, the correct option is B.

Note:
n-factor basically defines the amount of the reacting species. The n factor varies as per the reaction and nature of the substance. It is also defined as the valency of the ions in the aqueous solution.