Question

What is the concentration of Cu in a solution if a 25.00 mL aliquot reacted with an excess of KI that requires 15.64 mL of 39.94 mM Na2S2O3 solution to titrate the liberated iodine?

Answer 1

Cu²⁺ ions in the Cu solution react with I^- ions in KI and form Cu+ ions and I₂ gas.

2 Cu^₂₊ (aq) + 2 I^- (aq) ---------> 2 Cu⁺ (aq) + I₂ (g)

Liberated I₂ gas then reacts with Na₂S₂O₃ and gives S₄O₆^2- and I^- ions.

I₂ (g) + 2 S₂O^₃2- (aq) -----------> 2 I^- (aq) + S₄O₆^2- (aq)

According to the both reactions,

The mole ratios of S₂O^₃2- : I₂ : Cu²⁺ = 2 : 1 : 2

Number of Na2S2O3 moles reacted with I2 gas = ( 39.94 * 10^-3 * 10^{-3 *} 15.64) mol

= 6.25 * 10^-4 mol

So number of liberated I2 moles = ( 6.25 * 10^-4 mol /2 ) = 3.125 * 10^-4 mol

number of Cu²⁺ moles were the sample = 6.25 * 10^-4 mol

In 25 ml solution, there are 6.25 * 10^-4 Cu²⁺ moles.

Concentration of the Cu in the solution = (6.25 * 10^-4 mol / 25 ml ) * 1000 ml

[Cu^{2+ ]} = 0.025 M