Question

1.17 g of an impure sample of oxalic acid was dissolved and made up to 200 mL with water. 10 mL of this solution in acidic medium required 8.5 mL of a solution of potassium permanganate containing 3.16g/L of oxidation. Calculate the percentage purity of oxalic acid.

Answer 1

Hint: The purity of the sample is calculated by dividing the weight of the pure form of the sample to the weight of the impure form of the sample and then multiplying it with 100. To find the weight of the impure sample use the relation number of gram equivalent of the oxidizing agent is equal to the gram equivalent weight of the reducing agent.

Complete step by step solution:
The question is based on the redox reaction, so we can say the number of gram equivalent of the oxidizing agent is equal to the number of gram equivalent weight of the reducing agent.
No. of gram equivalent of oxidizing agent = No. of gram equivalent weight of reducing agent.
Or we can say, ${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$
Number of gram equivalent is equal to the mass of the solute divided by the equivalent mass of the solute.
$\text{No}\text{. of gram equivalent = }\dfrac{\text{mass of the solute}}{\text{equivalent mass of solute}}$
Let the weight of pure form of oxalic acid = x
The equivalent mass of oxalic acid = 63
$\text{No}\text{. of gram equivalent of oxalic acid = }\dfrac{\text{mass of the oxalic acid}}{\text{equivalent mass of oxalic acid}}=\dfrac{x}{63}$
$Normality=\dfrac{\text{number of gram equivalent}}{\text{volume of the solution in mL}}\text{ x 1000}$
So, the normality of oxalic acid = $\dfrac{x}{63}\text{ x }\dfrac{1000}{200}$
Volume of oxalic acid solution used = 10 mL
So, ${{N}_{1}}{{V}_{1}}=\dfrac{x\text{ x 1000}}{63\text{ x 200}}\text{ x 10}$
Now, for potassium permanganate,
$\text{No}\text{. of gram equivalent = }\dfrac{\text{mass of the solute}}{\text{equivalent mass of solute}}$
$\text{No}\text{. of gram equivalent of potassium permanganate = }\dfrac{\text{mass }}{\text{equivalent mass}}=\dfrac{3.16}{31.6}$
Number of gram equivalent of potassium permanganate is the normality because the mass is given in per liter form i.e., 3.16 g/L
Volume of potassium permanganate used = 8.5 mL
So, ${{N}_{2}}{{V}_{2}}=\dfrac{3.16}{31.6}\text{ x 8}\text{.5}$
So, equating both, we get
$\dfrac{x\text{ x 1000}}{63\text{ x 200}}\text{ x 10 = }\dfrac{3.16}{31.6}\text{ x 8}\text{.5}$
$x=1.071\text{ g}$
The weight of pure form of oxalic acid is 1.071 g.
The purity of the sample is calculated by dividing the weight of the pure form of the sample to the weight of the impure form of the sample and then multiplying it with 100.
Weight of impure oxalic acid given = 1.17 g
$\text{percentage purity = }\dfrac{\text{pure weight}}{\text{impure weight}}\text{ x 100}$
$\text{percentage purity = }\dfrac{\text{pure weight}}{\text{impure weight}}\text{ x 100 = }\dfrac{1.071}{1.17}\text{ x 100 = 91}\text{.53 }\!\!%\!\!\text{ }$

So, the purity percentage of oxalic acid is 91.53%.

Note: The equivalent mass of oxalic acid is calculated by dividing the molar mass to the basicity. The molar mass of oxalic acid is 126 and has basicity 2, hence the equivalent mass is 63. The equivalent mass of potassium permanganate is calculated by dividing the molar mass to the number of electrons gained or lost during the reaction. The molar mass of potassium permanganate is 158.04 and no. of electrons gained or lost is 5, hence the equivalent mass is 31.6.