An unknown Cu solution was prepared by dissolving 3.1373 g of the sample in a 200.0 mL flask. A 25.00 mL aliquot of this solution was titrated with 0.1084 M Na2S2O4, the titration required 15.93 mL to reach the end point. Calculate the % Cu in the sample?

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Michael Share
Feb 12, 2018

55.96 %

Explanation:

Dithionate(III) ions are reducing:

$\textsf{{\stackrel{\textcolor{red}{+ 3}}{S}}_{2} {O}_{4}^{2 -} + 2 {H}_{2} O \rightarrow 2 H \stackrel{\textcolor{red}{+ 4}}{S} {O}_{3}^{-} + 2 {H}^{+} + 2 e \text{ } \textcolor{red}{\left(1\right)}}$

The net oxidation no. change is $\textsf{+ 6 \rightarrow + 8}$ so 2 electrons are given out.

These are taken in by the copper(II):

$\textsf{C {u}^{2 +} + e \rightarrow C {u}^{+} \text{ } \textcolor{red}{\left(2\right)}}$

To get the electrons to balance we X sf(color(red)((2)) by 2 and add to $\textsf{\textcolor{red}{\left(1\right)} \Rightarrow}$

$\textsf{{S}_{2} {O}_{4}^{2 -} + 2 {H}_{2} O + 2 C {u}^{2 +} + \cancel{2 e} \rightarrow 2 H S {O}_{3}^{-} + 2 {H}^{+} + \cancel{2 e}}$

$\therefore$ 1 mole $\textsf{{S}_{2} {O}_{4}^{2 -} \equiv}$ 2 moles $\textsf{C {u}^{2 +}}$

$\textsf{c = \frac{n}{v}}$

$\therefore$$\textsf{n = c \times v}$

$\therefore$$\textsf{{n}_{{S}_{2} {O}_{4}^{2 -}} = 0.1084 \times \frac{15.93}{1000} = 0.0017268}$

$\therefore$$\textsf{{n}_{C {u}^{2 +}} = 0.0017268 \times 2 = 0.0034536}$

$\textsf{m = n \times {A}_{r}}$

$\therefore$$\textsf{{m}_{C u} = 0.0034536 \times 63.546 = 0.219464 \textcolor{w h i t e}{x} g}$

This is the mass of Cu(II) in 25.00 ml.

$\therefore$ 1 ml contains $\textsf{\frac{0.219464}{25.00} \textcolor{w h i t e}{x} g}$

$\therefore$ 200 ml contains $\textsf{\frac{0.219464}{25.00} \times 200.0 = 1.7557 \textcolor{w h i t e}{x} g}$

$\therefore$ the % of Cu(II) in the sample = $\textsf{\frac{1.7557}{3.1373} \times 100 = 55.96}$

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