Question

The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. 0.305 gm of a copper(II) salt was dissolved in water and added to an excess of potassium iodide solution liberating iodine according to the following equation.
$2C{u}^{2+}\left(aq\right)+4I^{-}\left(aq\right)\rightleftharpoons 2CuI\left(s\right)+I_2\left(aq\right)$
The iodine liberated required $24.5c{m}^3$ of a $0.100moled{m}^{-3}$ solution of sodium thiosulphate
$2S_2{O_3}^{2-}\left(aq\right)+I_2\left(aq\right)\to 2I^{-}\left(aq\right)+S_4{O_6}^{2-}\left(aq\right)$
the percentage of copper, by mass in the copper(II) salt is: $[Atomicmassofcopper=63.5]$

Answer 1

$2C{u}^{2+}+4I^{-}\rightleftharpoons 2CuI\left(s\right)+I_2$

$2S_2{O}_3^{2-}+I_2\rightleftharpoons 2I^{-}+S_4{O}_6^{2-}$

m moles of hypo$=$ m moles of iodine $\times 2$

$=$ m moles of $C{u}^{2+}$ ions.

$=24.5\times 0.1$ m moles.

So, mass of Copper$=\left(24.5\right)\left(0.1\right)\left({10}^{-3}\right)\left(63.5\right)$ gms.

So, $\%$ of Copper$=\frac{\left(24.5\right)\left({10}^{-3}\right)\left(0.1\right)\left(63.5\right)}{0.305}\times 100=51\%$