# A problem in titration

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1. Jun 11, 2017

### Wrichik Basu

1. The problem statement, all variables and given/known data:

The percentage of copper in a Copper (II) salt can be determined by using a thiosulphate titration. 0.305g of a copper (II) salt was dissolved in water and added to an excess of KI solution liberating Iodine. The liberated Iodine required 24.5 $dm^3$ of a 0.1 mole $dm^{-3}$ solution of sodium this sulphate. The percentage of copper, by mass, in the copper (II) salt is:

1. 64.2
2. 51.0
3. 48.4
4. 25.5

2. Relevant equations:

3. The attempt at a solution:

0.1 mole per $dm^3$ means 0.1 mole per litre.

Valence factor for sodium thiosulphate is 2.

So, there are 0.1×2=0.2 eqs of thiosulphate in 1 litre, which basically means the normality is 0.2.

No. of eqs of $Cu^{2+}$ = No. of eqs of Iodine liberated = No. of eqs of thiosulphate used = $\frac {24.5}{1000} ×0.2 = 4.9 ×10^{-3}$.

So, weight of $Cu^{2+}$ ions = $4.9 ×10^{-3} ×63.5=0.31115g$, which is greater than the weight of the sample.

Where am I going wrong?

2. Jun 11, 2017

### Staff: Mentor

Start by finding the equation of the reaction between iodine and thiosulfate.

3. Jun 12, 2017

### Wrichik Basu

The balanced equations are:
$$2Cu^{2+}+4I^{-} \rightarrow 2CuI+I_2$$
$$2S_2 O_3 ^{2-} +I_2 \rightarrow 2I^{-} +S_2 O_6^{2-}$$

4. Jun 12, 2017

### Staff: Mentor

5. Jun 12, 2017

### Wrichik Basu

Can't I do it by equivalents? Or does that concept fail here?

6. Jun 12, 2017

### Staff: Mentor

It doesn't fail when applied correctly.

7. Jun 12, 2017

### Wrichik Basu

Where am I going wrong?

8. Jun 12, 2017

### Staff: Mentor

Follow the stoichiometry and you will find out.

9. Jun 12, 2017

### Wrichik Basu

By stoichiometry I get answer 25.5.

But if I follow equivalents and do $4.9×10^{-3}×\dfrac {63.5}{2}$ (without reason why I divided by 2), I get 51.0, which is double the first answer.

Which is correct?

10. Jun 12, 2017

### Staff: Mentor

I told you - stoichiometry is right. Equivalents are a proxy for stoichiometry, and they can be quite convenient, but not when applied blindly. Your calculation of what the equivalent is was wrong (you calculated an equivalent for the reaction with H+, which is not what is happening here) so you got the wrong answer.