JEE MainChemistryRedox Reactions
A 6.35 g sample of a copper ore is dissolved and all its copper is converted to Cu ²⁺ ions. The solution is treated with an excess of KI , and the liberated iodine requires 40.0 mL of 1.0 M Na ₂ S ₂ O ₃ solution for complete titration. The percentage of copper in the ore sample is ____. (Given: Atomic mass of Cu = 63.5 u )
Correct answer
40
Step-by-step solution
The reactions involved are: 2 Cu ²⁺ + 4 I ^- Cu ₂ I ₂ + I ₂ I ₂ + 2 Na ₂ S ₂ O ₃ 2 NaI + Na ₂ S ₄ O ₆ From the principle of equivalence: Milliequivalents of Cu ²⁺ = Milliequivalents of I ₂ = Milliequivalents of Na ₂ S ₂ O ₃ The n-factor for Cu ²⁺ in this reaction is 1 (since it is reduced to Cu ^+ ). The n-factor for Na ₂ S ₂ O ₃ is 1 . Milliequivalents of Na ₂ S ₂ O ₃ = 40.0 1.0 1 = 40 Therefore, milliequivalents of Cu ²⁺ = 40 Moles of Cu = 40 1000 = 0.04 mol Mass of Cu = 0.04 63.5 = 2.54 g Percentage of Cu = ( 2.