Question

One way in which the useful metal copper is produced is by dissolving the mineral azurite, which contains copper(II) carbonate, in concentrated sulfuric acid. The sulfuric acid reacts with the copper(II) carbonate to produce a blue solution of copper(II) sulfate. Scrap iron is then added to this solution, and pure copper metal precipitates out because of the following chemical reaction: Fe(s) + CuSO4(aq) → Cu(s) + FeSO4(aq) Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 100.mL copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 52.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant digits.

Answer 1

Answer:

The original concentration of copper(II) sulfate in the sample is 0.0082 mol/L.

Explanation:

$Fe(s) + CuSO_4(aq)\rightarrow Cu(s) + FeSO_4(aq)$

Mass of copper precipitated out = 52 mg = 0.052 g

Moles of copper = $\frac{0.052 g}{63.5 g/mol}=0.0008189 mol$

According to reaction, 1 mole of copper is obtained from 1 mole of copper sulfate.

Then 0.0008189 mole of copper will be obtained from:

$\frac{1}{1}\times 0.0008189 mol=0.0008189 mol$ copper sulfate

Moles of copper sulfate = 0.0008189 mol

$CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}$

1 mole of copper sulfate produces 1 mole of copper (II) ions.Then 0.0008189 moles of copper sulfate will produce:

$1\times 0.0008189 mol=0.0008189 mol$ copper (II) ions.

$Concentration = \frac{Moles}{Volume (L)}$

Concentration of copper (II) ions = $[Cu^{2+}]$

Volume of the solution = 100 mL = 0.100 L

$[Cu^{2+}]=\frac{0.0008189 mol}{0.100 L}=0.008189 mol/L\approx 0.0082 mol/L$

The original concentration of copper(II) sulfate in the sample is 0.0082 mol/L.