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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Answer 1 · 2018-02-12T17:35:56

55.96 %

Start with the 1/2 equations:

Dithionate(III) ions are reducing:

$sf(stackrel(color(red)(+3))S_2O_4^(2-)+2H_2Orarr2Hstackrel(color(red)(+4))SO_3^(-)+2H^(+)+2e" "color(red)((1)))$

The net oxidation no. change is $sf(+6rarr+8)$ so 2 electrons are given out.

These are taken in by the copper(II):

$sf(Cu^(2+)+erarrCu^(+)" "color(red)((2)))$

To get the electrons to balance we X $sf(color(red)((2))$ by 2 and add to $sf(color(red)((1))rArr)$

$sf(S_2O_4^(2-)+2H_2O+2Cu^(2+)+cancel(2e)rarr2HSO_3^(-)+2H^(+)+cancel(2e))$

$:.$ 1 mole $sf(S_2O_4^(2-)-=)$ 2 moles $sf(Cu^(2+))$

$sf(c=n/v)$

$:.$ $sf(n=cxxv)$

$:.$ $sf(n_(S_2O_4^(2-))=0.1084xx15.93/1000=0.0017268)$

$:.$ $sf(n_(Cu^(2+))=0.0017268xx2=0.0034536)$

$sf(m=nxxA_r)$

$:.$ $sf(m_(Cu)=0.0034536xx63.546=0.219464color(white)(x)g)$

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

$:.$ 1 ml contains $sf(0.219464/(25.00)color(white)(x)g)$

$:.$ 200 ml contains $sf(0.219464/(25.00)xx200.0=1.7557color(white)(x)g)$

$:.$ the % of Cu(II) in the sample = $sf(1.7557/(3.1373)xx100=55.96)$