1. Tardigrade
  2. Question
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  4. The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. 0.305 g 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: 2 Cu 2+( aq )+4 I -( aq ) leftharpoons 2 CuI ( s )+ I 2( aq ) The iodine liberated required 24.5 cm 3 of a 0.100 mol dm -3 solution of sodium thiosulphate. 2 S 2 O 32-( aq )+ I 2( aq ) longrightarrow 2 I -( aq )+ S 4 O 62-( aq ) The percentage of copper (by mass) in the copper(II) salt is [Atomic mass of copper =63.5 ]

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Q. The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. $0.305\, g$ 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:
$2 Cu ^{2+}( aq )+4 I ^{-}( aq ) \rightleftharpoons 2 CuI ( s )+ I _{2}( aq )$
The iodine liberated required $24.5\, cm ^{3}$ of a $0.100 \,mol\,dm ^{-3}$ solution of sodium thiosulphate.
$2 S _{2} O _{3}^{2-}( aq )+ I _{2}( aq ) \longrightarrow 2 I ^{-}( aq )+ S _{4} O _{6}^{2-}( aq )$
The percentage of copper (by mass) in the copper(II) salt is [Atomic mass of copper $=63.5$ ]

Some Basic Concepts of Chemistry

Solution:

$2 \times$ mol of $I_{2}=$ Mol of hypo.
So mol of $I _{2}=\frac{24.5 \times 0.1 \times 10^{-3}}{2}$
Moles of $Cu ^{2+}=2 \times \frac{24.5 \times 0.1 \times 10^{-3}}{2}$