Question

A solution of Na2CO3 is added dropwise to a solution that contains 1.15×10−2 M Fe2+ and 0.58×10−2 M Cd2+. What concentration of CO32− is need to initiate precipitation of the first ion? Neglect any volume changes during the addition.

Answer 1

The question is incomplete, complete question is;

A solution of $Na_2CO_3$ is added dropwise to a solution that contains$1.15\times 10^{-2} M$ of $Fe^{2+}$ and $0.58\times 10^{-2} M$ and $Cd^{2+}$.

What concentration of $CO_3^{2-}$ is need to initiate precipitation? Neglect any volume changes during the addition.

$K_{sp}$ value $FeCO_3: 2.10\times 10^{-11}$

$K_{sp}$ value $CdCO_3: 1.80\times 10^{-14}$

What concentration of $CO_3^{2-}$ is need to initiate precipitation of the first ion.

Answer:

Cadmium carbonate will precipitate out first.

Concentration of $CO_3^{2-}$ is need to initiate precipitation of the cadmium (II) ion is $3.103\times 10^{-12} M$.

Explanation:

1) $FeCO_3\rightleftharpoons Fe^{2+}+CO_3^{2-}$

The expression of an solubility product of iron(II) carbonate :

$K_{sp}=[Fe^{2+}][CO_3^{2-}]$

$2.10\times 10^{-11}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{2.10\times 10^{-11}}{1.15\times 10^{-2} M}$

$[CO_3^{2-}]=1.826\times 10^{-9}M$

2) $CdCO_3\rightleftharpoons Cd^{2+}+CO_3^{2-}$

The expression of an solubility product of cadmium(II) carbonate :

$K_{sp}=[Cd^{2+}][CO_3^{2-}]$

$1.80\times 10^{-14}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{1.80\times 10^{-14}}{0.58\times 10^{-2} M}$

$[CO_3^{2-}]=3.103\times 10^{-12} M$

On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.

So, cadmium carbonate will precipitate out first.

And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the $3.103\times 10^{-12} M$ concentration.