Question

20 mL of 0.02 M hypo solution is used for the titration of 10 mL of copper sulphate solution, in the presence of excess of KI using starch as an indicator. The molarity of \(Cu^{2+}\) is found to be ________ × \(10^{–2} M\). [nearest integer]

Answer 1

Correct Answer: 4

To determine the molarity of \(Cu^{2+}\), we use the following titration reaction in the presence of excess KI:

\(2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2\)
\(2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-\)

The iodine formed in the first reaction reacts with the hypo (thiosulfate) solution in the second reaction. From the stoichiometry of the reactions:

1. 1 mole of \(I_2\) reacts with 2 moles of \(\text{S}_2\text{O}_3^{2-}\).
2. 1 mole of \(Cu^{2+}\) produces 0.5 moles of \(I_2\).

Given that 20 mL of 0.02 M \(\text{S}_2\text{O}_3^{2-}\) is used:

\(\text{Moles of S}_2\text{O}_3^{2-} = 0.02 \times 0.020 = 0.0004 \text{ moles}\)

The moles of \(\text{I}_2\) that reacted:

\(\text{Moles of I}_2 = \frac{0.0004}{2} = 0.0002 \text{ moles}\)

Moles of \(Cu^{2+}\) are therefore:

\(\text{Moles of } Cu^{2+} = 2 \times 0.0002 = 0.0004 \text{ moles}\)

Given the volume of the copper sulfate solution is 10 mL or 0.01 L, the molarity of \(Cu^{2+}\) is:

\(M = \frac{0.0004}{0.01} = 0.04 \text{ M}\)

Expressed in the format \(10^{-2}\) M, the molarity is \(4 \times 10^{-2}\) M.

This result, 4, lies within the expected range of 4 to 4. Thus, the final answer is 4.