Question

A mixture of $~{{H}_{2}}S{{O}_{4~}}$ and ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ oxalic acid) and some inert impurity weighing 3.185g was dissolved in water and the solution was made up to 1litre. 10mL of this solution required 3mL of 0.1N NaOH for complete neutralization. In another experiment, 100mL of the same solution in hot conditions requires 4mL of 0.02M $KMn{{O}_{4}}$ solution for complete reaction.
The weight % of $~{{H}_{2}}S{{O}_{4~}}$ in the mixture was:
A. 40
B. 50
C. 60
D. 80

Answer 1

. To calculate the weight percent of $~{{H}_{2}}S{{O}_{4~}}$ in the mixture, we should start by doing acid base conversion. And we should focus on the n factor in this question.

Complete step by step answer:
In the question, it is given that the mixture of sulphuric acid and oxalic acid is reacting with NaOH.
So, we should start by assuming milli-moles for $~{{H}_{2}}S{{O}_{4~}}$ and ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ for reaction with NaOH.

Let us assume that, x and y be the milli-moles of $~{{H}_{2}}S{{O}_{4~}}$ and ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ in given mixture when both reacted with base.
Now, for acid base reaction we should note that:
Milli-equivalent of acid = milli-equivalent of base
One important thing that we should focus on is the n factor or basicity of both acids. We should note that the factor or basicity of both acids is 2.

& x\times 2+y\times 2=3\times 0.1\times \frac{1000}{10} \\\ & x+y=15 \\\ \end{aligned}$$ Now, in the second part of question, it is given that we did a second experiment of reacting $KMn{{O}_{4}}$ with the same solution of acids. In second experiment, we should note that only $KMn{{O}_{4}}$ react with ${{H}_{2}}{{C}_{2}}{{O}_{4}}$. Ionic equation $$Mn{{O}^{-4}}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O$$ $M{{n}^{7+~}}$ is converted into $M{{n}^{2+}}$ $${{C}_{2}}{{O}_{4}}^{2-}\to 2C{{O}_{2}}+2{{e}^{-}}$$ ${{C}_{2}}{{O}_{4}}^{2-}$ converted into $C{{O}_{2}}$ Now, we will do similarly with the first part. Mili-equivalent of ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ = mili-equivalent of $KMn{{O}_{4}}$. $$\begin{aligned} & y\times 2 = 4\times 0.02\times 5\times \dfrac{1000}{100} \\\ & y=2 \\\ \end{aligned}$$ And from the first part: $$\begin{aligned} & x+y=15 \\\ & x=15-2 \\\ & x=13 \\\ \end{aligned}$$ So, now we will calculate the weight of $~{{H}_{2}}S{{O}_{4~}}$. Weight of $~{{H}_{2}}S{{O}_{4~}}$ = $13\times {{10}^{-3}}\times 98 = 1.274g$ Weight % of $~{{H}_{2}}S{{O}_{4~}}$ in sample = $\dfrac{1.274}{3.185}\times 100$ = 40% **So, the correct answer is “Option A”.** **Note:** We should note that when we dissolve solid in a solvent to create a solution, the concentration of the solution can be expressed in a variety of ways. We should note that the concept of moles and equivalents, and thus millimoles and milliequivalents, underlies this relationship. We should focus on that the equivalent unit was introduced to describe that when solutes dissolve in solvent to create a solution, the number of particles dispersed depends on the valence of the solute. For example, when one molecule of KCl dissolves, it leaves two ions, or charged particles − a ${{K}^{+}}$ ion and a $C{{l}^{-}}$ ion. This means that KCl has a valence of 2.