Question

A $0.276$ g impure sample of copper ore is dissolved and $C{u}^{2+}$ is titrated with $KI$ solution. $I_2$ liberated required $40$ mL of $0.1MN{a}_2{S}_2{O}_3$ solution for titration. The $\%$ of impurities in the ore is:

Answer 1

The balanced net ionic equation is as follows:

$C{u}^{2+}+2I^{-}\to Cu{I}_2$

Thus, $2N{a}_2{S}_2{O}_3=I_2=2C{u}^{2+}$.

We know, $2$ moles of $N{a}_2{S}_2{O}_3$ require $1$ mole of $I_2$.
So, number of milli-moles of $I_2$ liberated $=\frac{4}{2}=2$
According to equation, millimoles of $C{u}^{+2}$ used $=2\times 2=4$
So, mass of $C{u}^{+2}$ in solution $=4\times \frac{63.5}{1000}=0.24$

So, $\%$ of impurities $=(0.27-0.24)\times \frac{100}{0.279}=8\%$