A solution of 0.4 g of a compound containing C u 2 + and C 2 O 4 2 − ions on titration with 0.02 M - KMn O 4 in the presence of H 2 S O 4 consumes 67.8 ml of the oxidant. The resultant solut

Question

A solution of 0.4 g of a compound containing $C u^{2 +}$ and $C_{2} O_{4}^{2 -}$ ions on titration with 0.02 M - $KMn O_{4}$ in the presence of $H_{2} S O_{4}$ consumes 67.8 ml of the oxidant. The resultant solution is neutralized with $N a_{2} C O_{3}$ , acidified with dilute acetic acid and treated with excess $KI$ . The liberated iodine requires 22.6 ml of 0.05 M - $N a_{2} S_{2} O_{3}$ for complete reduction. If the molar ratio of $C u^{2 +}$ to $C_{2} O_{4}^{2 -}$ in the compound is 1:x, the value of $x$ is

Answer:

3.00

Solution:

$KMn O_{4}$ = 0.02 × 0.0678 = 1.356 mmol \equiv 5 × 1.356 = 6.78 meq, so $C_{2} O_{4}^{2 -}$ = 6.78/2 = 3.39 mmol. Thiosulphate = 0.05 × 0.0226 = 1.13 mmol = $C u^{2 +}$ . Ratio Cu : $C_{2} O_{4}$ = 1.13 : 3.39 = 1 : 3, so $x$ = 3. Ans: 3.

TestHub · Accepted Answer

KMn O 4 ​ = 0.02 × 0.0678 = 1.356 mmol \equiv 5 × 1.356 = 6.78 meq, so C 2 ​ O 4 2 − ​ = 6.78/2 = 3.39 mmol. Thiosulphate = 0.05 × 0.0226 = 1.13 mmol = C u 2 + . Ratio Cu : C 2 ​ O 4 ​ = 1.13 : 3.39 = 1 : 3, so x = 3. Ans: 3.

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