Question

100 ml $CuS{O}_4$ (aq) was electrolyzed using inert electrodes by passing 0.965 A till the $pH$ of the resulting solution was 1. The solution after electrolysis was neutralized, treated with excess $KI$ and titrated with 0.04 M $N{a}_2{S}_2{O}_3$. Volume of $N{a}_2{S}_2{O}_3$ required was 35 ml. Assume no volume change during electrolysis. If the initial concentration (M) of $CuS{O}_4$ was $64\times {10}^{-x}$, then what is the value of $x$?

Answer 1

$I_2+N{a}_2{S}_2{O}_3\to I^{-}+N{a}_2{S}_4{O}_6$

moles of $N{a}_2{S}_2{O}_3=0.04\times 35\times {10}^{-3}$

moles of $I_2=\frac{0.04\times 35\times {10}^{-3}}{2}moles$

moles of $KI=2\times \frac{0.04\times 35\times {10}^{-3}}{2}moles$

$KI$ and $CuS{O}_4$ reaction can be written as:

$4KI+2CuS{O}_4\to 2CuI+I_2+2K_{2}S{O}_4$

moles of $CuS{O}_4$ reacted with $KI=\frac{0.04\times 35\times {10}^{-3}}{2}=0.7\times {10}^{-3}moles$

Concentration of $CuS{O}_4$ reacted$=0.7\times {10}^{-3}\times 10=7\times {10}^{-3}M$

Concentration of $CuS{O}_4$ electrolyzed$=(64-7)\times {10}^{-3}=57\times {10}^{-3}M$

Therefore, $x=3$