Excess KI is added to a solution containing 0.00200 mol of Cu^2+ (2Cu^2+ + 4I- -> 2CuI + I2). The liberated iodine is titrated with 0.100 M Na2S2O3. The volume of thiosulfate required is:

NEET levelSingle correct optionDifficulty 1 of 3eq120

Excess KI is added to a solution containing 0.00200 mol of $Cu X^{2 +}$ ( $2 Cu X^{2 +} + 4 I X^{-} 2 CuI + I X_{2}$ ). The liberated iodine is titrated with 0.100 M $Na X_{2} S X_{2} O X_{3}$ . The volume of thiosulfate required is:

Two

Cu X^{2 +}

give one

I X_{2}

, and one

I X_{2}

consumes 2 thiosulfate, so 1 mol

Cu X^{2 +}

\equiv

1 mol thiosulfate. Moles thiosulfate = 0.00200 mol, volume = 0.00200/0.100 = 0.0200 L = 20.0 mL.