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Question

# A solution of 0.2 g of a compound containing Cu2+ and C2O2–4 ions on titration with 0.02 M KMnO4 in presence of H2SO4 consumes 22.6 mL of the oxidant. The resultant solution is neutralized with Na2CO3, acidified with dil. acetic acid and treated with excess KI.The liberated iodine requires 11.3 mL of 0.05MNa2S3O3 solution for complete reduction. Find out the mole ratio of Cu2+ to C2O2–4 in the compound is a:b. The sum of a+b is

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Solution

## In the given problem, a solution containing Cu2+ and C2O2−4 is titrated first with KMnO4 and then with Na2S2O3 in presence of KI. In titration with KMnO4, it is the C2O2−4 ions that react with the MnO−4 ions. The concerned balanced equation may be written as below. 2MnO−4+5C2O2−4+16H+→2Mn2++10CO2+8H2O Thus according to the above reaction 2 mol MnO−4 reacts with 5 mol of C2O2−4 However, No. of mmol of MnO−4 and used in titration =0.02×22.6 Now, milimoles of oxalate =0.02×22.6×52=1.130 KI reacts with Cu2+ as 2Cu2++4I−→Cu2I2+I2 I2+2S2O−3→2I−+S4O2−6 According to the above reactions mmoles of thiosulfate =mmoles of Cu2+ =0.05×11.3 ∴, Mole ratio of Cu2+ to C2O2−4=0.565 mmol1.130 mmol=1:2 Hence, a:b=1:2 So, a+b=1+2=3

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