Question

Concentrated sodium hydroxide can separate a mixture of:
(A) $Z{n^{2 + }}$ and $P{b^{2 + }}$
(B) $A{l^{3 + }}$ and $Z{n^{3 + }}$
(C) $C{r^{3 + }}$ and $F{e^{3 + }}$
(D) $A{l^{3 + }}$ and $C{r^{3 + }}$

Answer 1

Hint: As before we start the solution of the question firstly, we discuss some details about concentrated sodium hydroxide. Its special property is that it can form many types of hydrates and it is odourless. Whereas, when it is in its pure form it is a white crystalline solid and it has higher viscosity as compared to water.

Complete Step by Step Solution:
To know the separation of mixture firstly we take $C{r^{3 + }}$ and react it with conc. $NaOH$ (sodium hydroxide), from doing the following reaction, we get,
$C{r^{3 + }} + O{H^ - }\xrightarrow{{}}Cr{(OH)_3}$

As per the further reaction gives,
$Cr{(OH)_3} + O{H^ - }\xrightarrow{{}}Cr{(OH)_4}^ - $
We know that $Cr{(OH)_3}$ is a green coloured solution from which $Cr{(OH)_4}^ - $ will also be the solution of green coloured.

Now after reacting $C{r^{3 + }}$, we can react $F{e^{3 + }}$with the same and we get,
$F{e^{3 + }} + 3O{H^ - }\xrightarrow{{}}Fe{(OH)_3}$
As in the above reaction we get that the product of the reaction $Fe{(OH)_3}$ is a precipitate.

From which we can say that, $Cr{(OH)_4}^ - $ is in the form of solution and $Fe{(OH)_3}$ is in the form of precipitate which is not in a from of solution,
So, as we know that the precipitate is removed from the solution,

According to the question, we can say that, it can be separated from the separated.
As a result we can say that, $C{r^{3 + }}$ and $F{e^{3 + }}$ can be separated using concentrated $NaOH$ .
Hence, the correct option is (C).

Note: When we write a reaction it is necessary to check that the equation is equally balanced from both sides and we have to know that valency of each element from which we can easily justify how many bonds this element is going to be made.