Question

638g of $CuS{{O}_{4}}$ solution is titrated with excess of 0.2M KI solution. The liberated ${{I}_{2}}$ required 400ml of 1.0M $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$for complete reaction. The percentage purity of in the $CuS{{O}_{4}}$sample is:
A. 5%
B. 10%
C. 15%
D. 20%

Answer 1

Hint: Here, we can see that iodine titration is done. In this we will first write the reaction, then will calculate the weight of $CuS{{O}_{4}}$and percentage of $CuS{{O}_{4}}$, by the formula:$\dfrac{{{w}_{CuS{{O}_{4}}}}}{{{w}_{solution}}}\times 100$

Complete answer:
In the classes of chemistry, we have carried out the laboratory experiment which deals with the basic experiment that is the titration.
We shall see in detail about the titration of the above given question.
- Iodometry is the direct titration of ${{I}_{2}}$ with the reducing agent. If the iodine is liberated by oxidising agent as${{I}^{-}}$ ion in acidic or basic media then this liberated ${{I}_{2}}$is titrated with reducing agent.
- We are given the values: 638g of$CuS{{O}_{4}}$ , 0.2M KI solution, 400ml of 1.0M $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$. We need to find out the percentage purity of $CuS{{O}_{4}}$ in the sample. So, we will write the reaction for this as:
\[CuS{{O}_{4}}+KI\to {{I}_{2}}+{{S}_{2}}{{O}_{3}}^{2-}\]
- The overall reaction can be written as:
\[\begin{align}
  & 2C{{u}^{2+}}+2{{e}^{-}}\to C{{u}_{2}}^{2+} \\
 & 2{{I}^{-}}\to {{I}_{2}}+2{{e}^{-}} \\
 & ----------- \\
 & 2CuS{{O}_{4}}+4KI\to C{{u}_{2}}{{F}_{2}}+2{{K}_{2}}S{{O}_{4}}+{{I}_{2}} \\
\end{align}\]
Now, the reduction of ${{I}_{2}}$is given by,
\[{{I}_{2}}+2{{S}_{2}}O_{3}^{2-}\to 2{{I}^{-}}+{{S}_{4}}O_{4}^{2-}\]
Here, we can see that 2 mol of $CuS{{O}_{4}}$= 4 mol KI = 1 mol${{I}_{2}}$ =2 mol${{S}_{2}}{{O}_{3}}^{2-}$
We can say that from 2 mol of $CuS{{O}_{4}}$,2mol${{S}_{2}}{{O}_{3}}^{2-}$ is produced.
This means that the mill moles of both are the same.
So, we can write that: milli moles of $CuS{{O}_{4}}$= milli moles ${{S}_{2}}{{O}_{3}}^{2-}$
= 400 $\times $ 1 = 400 milli mole =04 moles
Weight of 1 mol of $CuS{{O}_{4}}$ is 159.5g , we get it as:
\[63.546+32+4\times 16=159.5g\]
So, weight of $CuS{{O}_{4}}$= $0.4\times 159.5g=63.8g$
Now, percentage of $CuS{{O}_{4}}$ = $\dfrac{{{w}_{CuS{{O}_{4}}}}}{{{w}_{solution}}}\times 100$
\[\Rightarrow \dfrac{63.8}{638}\times 100=10\]

Therefore, we can conclude that the correct option is (B) that is the percentage purity of$CuS{{O}_{4}}$ in the sample is 10%.

Note:
Note that while calculating the percent purity, if an impure sample of a chemical of known percent purity is used in the chemical reaction, then the percent purity should be used in stoichiometric calculations.