An impure sample of sodium oxalate ( Na 2 C 2 O 4 ) weighing 0.20 g is dissolved in aqueous solution of H 2 SO 4 and solution is titrated at 7 0 ∘ C , requiring 45 mL of 0.02 M KMnO 4 solu

Question

An impure sample of sodium oxalate ( Na 2 ​ C 2 ​ O 4 ​ ) weighing 0.20 g is dissolved in aqueous solution of H 2 ​ SO 4 ​ and solution is titrated at 7 0 ∘ C , requiring 45 mL of 0.02 M KMnO 4 ​ solution. The end point is overrun, and back titration in carried out with 10 mL of 0.1 M oxalic acid solution. Find the % purity of Na 2 ​ C 2 ​ O 4 ​ in sample :

TestHub · Accepted Answer

Concepts: Redox titration, back titration, stoichiometry.

Moles of $K M n O_{4}$ used = $0.045 L \times 0.02 M = 0.0009 m o l$

Moles of oxalic acid used in back titration = $0.01 L \times 0.1 M = 0.001 m o l$

Moles of $N a_{2} C_{2} O_{4}$ reacted with $K M n O_{4}$ :

$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$

Moles of $K M n O_{4}$ reacted with oxalic acid = $\frac{2}{5} \times 0.001 = 0.0004 m o l$

Moles of $K M n O_{4}$ reacted with $N a_{2} C_{2} O_{4}$ = $0.0009 - 0.0004 = 0.0005 m o l$

Moles of $N a_{2} C_{2} O_{4}$ in the sample = $\frac{5}{2} \times 0.0005 = 0.00125 m o l$

Mass of $N a_{2} C_{2} O_{4}$ = $0.00125 m o l \times 134 g / m o l = 0.1675 g$

% purity = $\frac{0.1675}{0.20} \times 100 = 83.75%$

Chemistry - Redox Reactions Question with Solution | TestHub

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