The percentage of copper in copper (II) salt can b determined by using a thiosul phate titration. A 0.305 g of copper (II) salt was dissolved in water and added to an excess of potassium iodide solution liberating iodine according to the following equation:

$2{\mathrm{Cu}}^{2+}\left(aq\right)+4{\mathrm{I}}^{-}\left(aq\right)⟶2\mathrm{CuI}\left(s\right)+{\mathrm{I}}_2\left(aq\right)$

The iodine liberated required $24.5{\mathrm{cm}}^3$ of a 0.1 mole ${\mathrm{dm}}^{-3}$ solution of sodium thiosulphate for titration according to reaction :

$2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}\left(aq\right)+{\mathrm{I}}_2\left(aq\right)⟶2{\mathrm{I}}^{-}\left(aq\right)+{\mathrm{S}}_4{\mathrm{O}}_6^{2-}\left(aq\right)$

The percentage of copper by mass in the copper (II)

salt is: (Atomic mass of copper= 63.5