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 $\mathrm{Cu}^{2+}$ is found to be \_\_\_\_ $\times 10^{-2} \mathrm{M}$ [nearest integer] Given : $2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{I}_{2}$ $\mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}$

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Practical and Analytical Chemistry

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
____

\times 1 0^{- 2} M

[nearest integer]

Given :

2 Cu^{2 +} + 4 I^{-} \to Cu_{2} I_{2} + I_{2}

I_{2} + 2 S_{2} O_{3}^{2 -} \to 2 I^{-} + S_{4} O_{6}^{2 -}

Answer: 4

Solution

From the second reaction:

n_{eq}

I_{2}

n_{eq}

S_{2} O_{3}^{2 -}

n_{eq}

S_{2} O_{3}^{2 -}

20 \times 0.02 \times 1 = 0.4

meq

Since

I_{2}

has n-factor = 2:

n_{mol}

I_{2}

\frac{0.4}{2} = 0.2

mmol

From the first reaction: 2 moles of

Cu^{2 +}

produce 1 mole of

I_{2}

n_{mol}

Cu^{2 +}

2 \times 0.2 = 0.4

mmol

Molarity of

Cu^{2 +}

\frac{0.4}{10} = 0.04

M =

4 \times 1 0^{- 2}

M

Answer: 4

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Practical and Analytical Chemistry

Solution

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