Question

If $20\text{ }ml$ of $0.1M$${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is required to titrate $10\text{ }ml$ of a liquid iron supplement then the concentration of iron in the vitamin solution is:
(a) $1.2M$
(b) $2.4M$
(c) $0.6M$
(d) $1.56M$

Answer 1

Hint: It is a redox reaction which involves the gain and loss of the electrons and first we have to find the normality by equalizing the milliequivalents of both as $milliequivalents\text{ }of\text{ }C{{r}_{2}}O_{7}^{2-}=milliequivalents\text{ o}f\text{ }F{{e}^{3+}}$ and then from here we can find the normality of $F{{e}^{+3}}$ and then by applying the molarity formula as: $Molarity=\dfrac{normality}{no.\text{ of electrons involved}}$ we can easily find the concentration of the iron in the vitamin solution. Now solve it.

Complete step by step solution:
When potassium dichromate is made to react with the iron supplement, it undergoes redox reaction (i.e. those reactions in which one is oxidized at the cost of the other which undergoes reduction). The chromate ion undergoes reduction i.e. it gains the electrons whereas the iron undergoes oxidation i.e. it loses electrons. The redox reaction occurs as;
 $C{{r}_{2}}O_{7}^{2-}+F{{e}^{2+}}+14{{H}^{+}}\to 2C{{r}^{3+}}+F{{e}^{3+}}+7{{H}_{2}}O$
So, from this reaction, we can see that the chromium gains electrons and changes its oxidation state. The reaction occurs as;
$6{{e}^{-}}+C{{r}_{2}}O_{7}^{2-}\to 2C{{r}^{3+}}$ (1)
And iron loses electrons and undergoes change in the oxidation state. The reaction is supposed to occur as;
$F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{-}}$ (2)
Now, since it is a redox reaction, then ;
$milliequivalents\text{ }of\text{ }C{{r}_{2}}O_{7}^{2-}=milliequivalents\text{ o}f\text{ }F{{e}^{3+}}$ (3)
$milliequivalents\text{ }of\text{ }C{{r}_{2}}O_{7}^{2-}=volume\times normality\times no.\text{ }of\text{ }electrons\text{ }involved$
Volume of $C{{r}_{2}}O_{7}^{2-}=20\text{ }ml$(given)
Normality of $C{{r}_{2}}O_{7}^{2-}=0.1$(given)
from the equation (1), we can see that the number of electrons involved are 6. Then;
 $milliequivalents\text{ }of\text{ }C{{r}_{2}}O_{7}^{2-}=20\times 0.1\times 6=12$
Similarly, $milliequivalents\text{ o}f\text{ }F{{e}^{3+}}=10\times N\times 1$
Now, put these values in equation (3), we get;
$\begin{align}
& 12=10N \\
& N=1.2 \\
\end{align}$
Applying the molarity formula as, we can find the concentration of iron as;
$Molarity=\dfrac{normality}{no.\text{ of electrons involved}}$
As we know that :
$\begin{align}
& normality=1.2 \\
& no.\text{ }of\text{ }electrons\ involved=1\text{ }\left( given \right) \\
\end{align}$
The,
$Molarity=\dfrac{1.2}{1}=1.2M$
Hence, when$20\text{ }ml$ of $0.1M$${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is titrated with $10\text{ }ml$ of a liquid iron supplement then the concentration of iron in the vitamin solution is $1.2M$.

So, option (a) is correct.

Note: Don’t get confused in the words molarity and normality. By the molarity we mean the no. of moles of the solute to the total volume of the solution in liters . On the other hand, by the term normality, we mean the no. of moles of gram equivalent of the solute to total volume of the solution in liters.