Answer:
%Copper = 25.42 %
Explanation:
For solving this problem one first write down the half reactions as;
Reduction of Cu²⁺ to Cu⁺¹ by I⁻:
2 Cu²⁺ + 4 I⁻ → 2 CuI + I₂
Secondly, the titration reaction between Na₂S₂O₃ with I₂ as;
2 S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2 I⁻
Now, Calculate Moles of S₂O₃²⁻ as;
Moles = Molarity × Volume
Putting values,
Moles = 0.1 mol/L × 0.02 L
Moles = 0.002 moles of S₂O₃²⁻
From first two equations we can conclude that,
2 moles of S₂O₃²⁻ liberated I₂ produced by = 2 moles of Cu²⁺
So,
0.002 moles of S₂O₃²⁻ will liberate I₂ produced by = X moles of Cu₂⁺
Solving for X,
X = 0.002 mol × 2 mol / 2 mol
X = 0.002 mol of Cu²⁺
So, we require 0.002 moles of Cu²⁺ to produce I₂ enough for titrating with 0.1 M (20.00 mL) sodium thiosulphate solution.
Now, Calculate mass of Copper as,
Mass = Moles × A.Mass
Mass = 0.002 mol × 63.55 g/mol
Mass = 0.1271 g of Copper
Lastly, calculate % copper as,
%Copper = 0.1271 g / 0.500 g × 100
%Copper = 25.42 %
