100 mL of aqueous solution of 0.05 M Cu^{2+ | vedclass

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(C) The reaction between $Cu^{2+}$ and $I^-$ is: $2Cu^{2+} + 4I^- ightarrow 2CuI(s) + I_2$. Moles of $Cu^{2+} = 0.1 \ L 	imes 0.05 \ M = 0.005 \ mo

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(C) The reaction between $Cu^{2+}$ and $I^-$ is: $2Cu^{2+} + 4I^- ightarrow 2CuI(s) + I_2$. Moles of $Cu^{2+} = 0.1 \ L 	imes 0.05 \ M = 0.005 \ mol$. Moles of $I_2$ produced = $\frac{1}{2} 	imes 	ext{moles of } Cu^{2+} = 0.0025 \ mol$. The titration reaction is: $I_2 + 2Na_2S_2O_3 ightarrow 2NaI + Na_2S_4O_6$. Moles of $Na_2S_2O_3$ required = $2 	imes 	ext{moles of } I_2 = 2 	imes 0.0025 = 0.005 \ mol$. Volume of $Na_2S_2O_3 = \frac{	ext{moles}}{	ext{molarity}} = \frac{0.005 \ mol}{0.01 \ M} = 0.5 \ L = 500 \ mL$.

$100 \ mL$ of aqueous solution of $0.05 \ M$ $Cu^{2+}$ is added to $1 \ L$ of $0.1 \ M$ $KI$ solution. The resultant solution was titrated with $0.01 \ M$ $Na_2S_2O_3$ solution using starch indicator till blue color disappeared. What is the volume (in $mL$) of $Na_2S_2O_3$ used?

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