JEE MAIN - Chemistry (2022 - 26th July Evening Shift - No. 16)
$$20 \mathrm{~mL}$$ of $$0.02\, \mathrm{M}$$ hypo solution is used for the titration of $$10 \mathrm{~mL}$$ 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 : $$2 \,\mathrm{Cu}^{2+}+4 \,\mathrm{I}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{I}_{2}$$
$$ \mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-} $$
Answer
4
Explanation
$2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{l}_{2}+\mathrm{I}_{2}$
$\mathrm{I}_{2}+\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}$
Milliequivalents of hypo solution $=0.02 \times 20=0.4$
Milliequivalents of $\mathrm{Cu}^{2+}$ in $10 \mathrm{~mL}$ solution =
Milliequivalents of $\mathrm{I}_{2}=$ Milliequivalents of hypo = 0.4
Millimoles of $\mathrm{Cu}^{2+}$ ions in $10 \mathrm{~mL}=0.4$
Molarity of $\mathrm{Cu}^{2+}$ ions $=\frac{0.4}{10}=0.04 \,\mathrm{M}$
$$ =4 \times 10^{-2} \,\mathrm{M} $$
$\mathrm{I}_{2}+\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}$
Milliequivalents of hypo solution $=0.02 \times 20=0.4$
Milliequivalents of $\mathrm{Cu}^{2+}$ in $10 \mathrm{~mL}$ solution =
Milliequivalents of $\mathrm{I}_{2}=$ Milliequivalents of hypo = 0.4
Millimoles of $\mathrm{Cu}^{2+}$ ions in $10 \mathrm{~mL}=0.4$
Molarity of $\mathrm{Cu}^{2+}$ ions $=\frac{0.4}{10}=0.04 \,\mathrm{M}$
$$ =4 \times 10^{-2} \,\mathrm{M} $$
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