The percentage of copper in a copper (II) salt can be determined by using a thiosulfate titration. 0.305 g m of a copper (II) salt was dissolved in water, and an excess of potassium iodide was added t

Question

The percentage of copper in a copper (II) salt can be determined by using a thiosulfate titration. $0.305 g m$ of a copper (II) salt was dissolved in water, and an excess of potassium iodide was added to it, liberating iodine according to the following equation:

$2 C u^{2 +} (a q) + 4 I^{--} (a q) ⟶ 2 C u I (s) + I_{2} (a q)$

The iodine liberated required $24.5 c m^{3}$ of a $0.100 m o l d m^{-- 3}$ solution of sodium thiosulfate:

$\displaylines 2 S_{2} O_{3}^{2 -} (a q) + I_{2} (a q) ⟶ 2 I^{-} (a q) + S_{4} O_{6}^{2 -} (a q)$

The percentage of copper, by mass, in the copper (II) salt is:

TestHub · Accepted Answer

m moles of hypo = m moles of I₂ × 2 = m moles of Cu²⁺ ions = $24.5 \times 0.1$ m moles.

So, mass of copper = $24.5 \times 0.1 \times 1 0^{- 3} \times 63.5$ gm.

Thus, the percentage of copper is:

$% o f co pp er = \frac{24.5 \times 0.1 \times 1 0 ^{- 3} \times 63.5}{0.305} \times 100 \approx 51%$

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