Question

0.5g of fuming ${{{H}}_2}{{S}}{{{O}}_4}$ (oleum) is diluted with water. This solution is completely neutralized by ${{26}}{{.7mL}}$ of ${{0}}{{.4N}}$ ${{NaOH}}$. Find the percentage of free ${{S}}{{{O}}_3}$ in the sample solution.

Answer 1

Hint: This problem can be solved on the basis of mole concept. Before we calculate the percentage of free sulfur trioxide, we have to calculate three units of measurement, i.e. equivalent weight, total number of equivalents.

Complete step by step answer:
Oleum is also known as fuming sulfuric acid, i.e. it consists of sulfuric acid and free ${{S}}{{{O}}_3}$.
It is given that the amount of oleum, ${{{m}}_{{{ol}}}} = 0.5{{g}}$
Let ${{xg}}$ be the mass of ${{S}}{{{O}}_3}$, ${{{m}}_{{{S}}{{{O}}_3}}}$ in the oleum. So we can say that the mass of sulfuric acid, ${{{m}}_{{{{H}}_2}{{S}}{{{O}}_4}}} = \left( {0.5 - {{x}}} \right){{g}}$
The molecular weight of ${{S}}{{{O}}_3}$, ${{{M}}_{{{S}}{{{O}}_3}}} = 80{{g}}.{{mo}}{{{l}}^{ - 1}}$
From the value of molecular weight, we can calculate the value of equivalent weight of ${{S}}{{{O}}_3}$.
i.e. Equivalent weight of ${{S}}{{{O}}_3}$, ${{{E}}_{{{S}}{{{O}}_3}}} = \dfrac{{{{{M}}_{{{S}}{{{O}}_3}}}}}{{{V}}}$, ${{V}}$ is the valency, i.e. ${{V = 2}}$.
i.e. ${{{E}}_{{{S}}{{{O}}_3}}} = \dfrac{{80}}{{{2}}} = 40$
Similarly, we can calculate the equivalent weight of ${{{H}}_2}{{S}}{{{O}}_4}$.
Molecular weight of ${{{H}}_2}{{S}}{{{O}}_4}$, ${{{M}}_{{{{H}}_2}{{S}}{{{O}}_4}}} = 98{{g}}.{{mo}}{{{l}}^{ - 1}}$
So equivalent weight of ${{{H}}_2}{{S}}{{{O}}_4}$, ${{{E}}_{{{{H}}_2}{{S}}{{{O}}_4}}} = \dfrac{{{{{M}}_{{{{H}}_2}{{S}}{{{O}}_4}}}}}{{{V}}} = \dfrac{{98}}{2} = 49$
Now we can calculate the number of gram equivalents of ${{S}}{{{O}}_3}$, ${{{N}}_{{{S}}{{{O}}_3}}} = \dfrac{{{{{m}}_{{{S}}{{{O}}_3}}}}}{{{{{E}}_{{{S}}{{{O}}_3}}}}} = \dfrac{{{x}}}{{40}}$
Similarly, number of gram equivalents of ${{{H}}_2}{{S}}{{{O}}_4}$, ${{{N}}_{{{{H}}_2}{{S}}{{{O}}_4}}} = \dfrac{{{{{m}}_{{{{H}}_2}{{S}}{{{O}}_4}}}}}{{{{{E}}_{{{{H}}_2}{{S}}{{{O}}_4}}}}} = \dfrac{{{{0}}{{.5 - x}}}}{{49}}$
The total number of gram equivalents is the sum of number of gram equivalents of ${{S}}{{{O}}_3}$ and ${{{H}}_2}{{S}}{{{O}}_4}$.
Total number of gram equivalents, ${{{N}}_{{T}}} = \dfrac{{{x}}}{{40}} + \dfrac{{0.5 - {{x}}}}{{49}}$
Normality of ${{NaOH}}$, ${{{N}}_{{{NaOH}}}} = 0.4{{N = 0}}{{.4eq}}{{.}}{{{L}}^{ - 1}}$
${{1000mL}}$ will have $0.4{{eq}}$ of ${{NaOH}}$. So ${{26}}{{.7mL}}$ will have $\dfrac{{0.4}}{{1000}} \times 26.7 = 0.0107{{eq}}$
At equivalence point, the number of gram equivalents of ${{NaOH}}$ will be equal to the number of gram equivalents of ${{{H}}_2}{{S}}{{{O}}_4}$.
So $\dfrac{{0.4}}{{1000}} \times 26.7{{ = }}\dfrac{{{x}}}{{40}} + \dfrac{{0.5 - {{x}}}}{{49}}$
On simplification, we get
$0.0107 = \dfrac{{49{{x}} + 20 - 40{{x}}}}{{1960}} = \dfrac{{9{{x}} + 20}}{{1960}}$
On further simplification,
$0.0107 \times 1960 = 9{{x}} + 20 \Leftrightarrow 0.972 = 9{{x}} \Leftrightarrow {{x = 0}}{{.108}}$
Thus we get the mass of ${{S}}{{{O}}_3}$, \[{{{m}}_{{{S}}{{{O}}_3}}} = 0.108{{g}}\]
Now we can calculate the percentage of free ${{S}}{{{O}}_3}$, $\% = \dfrac{{{{{m}}_{{{S}}{{{O}}_3}}}}}{{{{{m}}_{{{ol}}}}}} \times 100 = \dfrac{{0.108}}{{0.5}} \times 100 = 21.6\% $

Thus the percentage of free ${{S}}{{{O}}_3}$ is $21.6\% $.

Note: We have to concentrate the steps given properly. There is a chance to get confused in calculating the percentage. Since we need to calculate the percentage of free ${{S}}{{{O}}_3}$, the equivalents of oleum will be different when we consider it in terms of mass.