Question

The solubility of $A{g_2}C{O_3}$ in water is $1.26 \times {10^{ - 4}}mol{\text{ litr}}{{\text{e}}^{ - 1}}$ . What is its solubility in $0.02M$ $N{a_2}C{O_3}$ solution?
Assume no hydrolysis of $C{O_3}^{2 - }$ ion. Take $^3\sqrt 2 = 1.26$

Answer 1

Hint : The solubility of the compound is given by the interaction of the ions of the molecule with the ions or molecules of the water. The ionic compounds are more soluble in polar solvents like water. Whereas ionic compounds are less soluble in those compounds which contain common ions or salts. Because of the less attraction or force of attraction with each other.

Complete Step By Step Answer:
The above question demands the solubility of the $N{a_2}C{O_3}$ in $0.02M$ concentration.
So let’s proceed with the question following each and every step, so let’s begin.
First we have to find the ‘ ${K_{sp}}$ ’ of the equation of the decomposition of the $A{g_2}C{O_3}$ in its ions.
So, the balanced equation of the decomposition of the $A{g_2}C{O_3}$ is:
$A{g_2}C{O_3} \to 2A{g^ + } + C{O_3}^{2 - } \\\ s{\text{ 2s s}} \\\$
Now, the ${K_{sp}}$ of the above equation will be:
${K_{sp}} = {\left[ {A{g^ + }} \right]^2}\left[ {C{O_3}^{2 - }} \right] \\\ {K_{sp}} = 4{s^3}{\text{ }} \\\$
Now let’s proceed to find the solubility:
Now let’s suppose that in $0.02M$ $A{g_2}C{O_3}$ , the solubility will be $'\lambda '$
Therefore, let’s write it correctly for the smooth going of the equation.
$\left[ {A{g^ + }} \right] = 2\lambda {\text{ }}\left[ {C{O_3}^{2 - }} \right] = 0.02$
Again writing the equation of ${K_{sp}}$
${K_{sp}} = {\left[ {A{g^ + }} \right]^2}\left[ {C{O_3}^{2 - }} \right]{\text{ }}$
$4{s^3} =$ ${\left( {2\lambda {\text{ }}} \right)^2}$ $\left( {0.02} \right)$
Since, we are given the value of sin the question:
$4{\left( {1.26 \times {{10}^{ - 4}}} \right)^3}$ $= {\left( {2\lambda {\text{ }}} \right)^2}$ $\left( {0.02} \right)$
On solving the equation and putting the values given in question, we will get:
$2 \times {10^{ - 12}} = {\lambda ^2} \times 2 \times {10^{ - 2}} \\\ {10^{ - 10}} = {\lambda ^2} \\\$
Hence our, $\lambda = {10^{ - 5}}$ $mol{\text{ litr}}{{\text{e}}^{ - 1}}$
Therefore the solubility of $A{g_2}C{O_3}$ in $0.02M$ $N{a_2}C{O_3}$ solution is $\lambda = {10^{ - 5}}$ $mol{\text{ litr}}{{\text{e}}^{ - 1}}$
The above procedure is quite tricky, but we just have to focus on the solubility constant of the equation and the ions or stoichiometric coefficient of the ions of the balanced equation.

Note :
Sodium Carbonate is the inorganic compound with the formula $N{a_2}C{O_3}$ and its various hydrates. All forms are white, odourless, water-soluble salts that yield moderately alkaline solutions in water. Historically, it was extracted from the ashes of plants growing in sodium-rich soils.