Question

If $\text{ }{{\text{K}}_{\text{sp}}}$ of $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$is $\text{ 8 }\times \text{1}{{\text{0}}^{-12}}\,\text{ }$, the molar solubility of $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$in $\text{ 0}\text{.1 M AgN}{{\text{O}}_{\text{3}}}$ is :
A) $\text{ 8}\times \text{1}{{\text{0}}^{-12}}\text{M}$
B) $\text{ 8}\times \text{1}{{\text{0}}^{-10}}\text{M}$
C) $\text{ 8}\times \text{1}{{\text{0}}^{-11}}\text{M}$
D) $\text{ 8}\times \text{1}{{\text{0}}^{-13}}\text{M}$

Answer 1

Hint: The molar solubility is the number of moles of a substance that can be dissolved per litre of the solution after achieving saturation. The molar solubility for salt $\text{ AxBy }$ is written as,
$\text{ AxBy }\rightleftharpoons \text{ x}{{\text{A}}^{\text{+}}}\text{ (aq) + y}{{\text{B}}^{-}}\text{(aq) }$
Solubility product is,
$\text{ }{{\text{K}}_{\text{sp}}}\text{ = }{{\left[ {{\text{A}}^{\text{+}}} \right]}^{\text{x}}}{{\left[ {{\text{B}}^{-}} \right]}^{\text{y}}}\text{ }$
The presence of the same ion in the solution reduces the solubility of the salt. The increased concentration of ions is taken into consideration while solving for the molar solubility.

Complete step by step answer:
We have given the following data on the problem.
Solubility product of silver carbonate$\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$ is $\text{ }{{\text{K}}_{\text{sp}}}\text{ = 8 }\times \text{1}{{\text{0}}^{-12}}\,\text{ }$
The $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$is in the $\text{ 0}\text{.1 M AgN}{{\text{O}}_{\text{3}}}$ solution.
We are interested to find out the molar solubility of $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$.
The solute like silver carbonate dissolves in the solution and forms corresponding silver and carbonate ions. The equilibrium reaction between the ions and solute is as shown below,
$\text{ }\begin{matrix}
{} & \text{A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{(s)} & \rightleftharpoons & \text{2A}{{\text{g}}^{\text{+}}}\text{(aq)} & \text{+} & \text{CO}_{\text{3}}^{\text{2-}} \\
\text{Before} & \text{S} & {} & \text{0} & {} & \text{0} \\
\text{After} & \text{-} & {} & \text{2S+0}\text{.1} & {} & \text{S} \\
\end{matrix}$
This $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$is dissolved in the $\text{ 0}\text{.1 M AgN}{{\text{O}}_{\text{3}}}$ solution. Both salts have the common$\text{ A}{{\text{g}}^{+}}\text{ }$. Thus, the silver ion from both the salt contributes towards the molar solubility.The solubility product for the $\text{ A}{{\text{g}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\text{ }$is written as follows,
$\text{ }{{\text{K}}_{\text{sp}}}\text{ = }{{\left[ \text{A}{{\text{g}}^{\text{+}}} \right]}^{2+}}\left[ \text{CO}_{3}^{2-} \right]\text{ }$
Let’s substitute the values in the above equation we have,
$\text{ 8}\times \text{1}{{\text{0}}^{-12}}\text{ = }{{\left( 0.1+2S \right)}^{2+}}\left( S \right)$
The value of solubility product is very small, thus we neglect it $\text{ 0}\text{.1 + 2S = 0}\text{.1 }$.The equation now becomes,
$\begin{align}
& \text{ 8}\times \text{1}{{\text{0}}^{-12}}\text{ = }{{\left( 0.1 \right)}^{2}}\left( S \right) \\
& S=\dfrac{\text{8}\times \text{1}{{\text{0}}^{-12}}}{{{\left( 0.1 \right)}^{2}}}=\text{8}\times \text{1}{{\text{0}}^{-10}}\text{M} \\
\end{align}$
Therefore, molar solubility of a silver carbonate in presence of silver nitrate is equal to $\text{ 8}\times \text{1}{{\text{0}}^{-10}}\text{M}$ .

Hence, (B) is the correct option.

Note: The common ion effect is the addition of the salt to a solution such that the two slats have the common ion.for example, silver carbonate and silver nitrate. Both salts have the common ion as silver. The solubility of the salt is measuredly affected by the common ion. The silver carbonate is less soluble as compared to silver nitrate, but the presence of silver nitrate increases the molar solubility of less soluble salt.