Question

Concentration of the $A{{g}^{+}}$ ions in a saturated solution of $A{{g}_{2}}{{C}_{2}}{{O}_{4}}$ is $2.2\times {{10}^{-4}}mol{{L}^{-1}}$. Solubility product of $A{{g}_{2}}{{C}_{2}}{{O}_{4}}$ is __________.
A. $5.3\times {{10}^{-12}}$
B. $2.42\times {{10}^{-8}}$
C. $2.66\times {{10}^{-12}}$
D. $4.5\times {{10}^{-11}}$

Answer 1

Hint: Solubility product is a constant or equilibrium constant for the dissociation of solid substances into aqueous solution. It depends on temperature and is represented by Ksp.

Complete step by step answer:
The property of the body by which the solute dissolves in a solvent to form a solution is called solubility.
Solubility product is the equilibrium constant which increases with increase in temperature since solubility increases with temperature.
Solubility product also depends upon lattice energy of salt and salvation enthalpy of ions. Most salts dissociate into ions when they dissolve.
Solubility is different for different salts:
Soluble Solubility$>0.1M$
Slightly soluble $0.01M<$ Solubility$<0.1M$
 Sparingly soluble Solubility $<0.1M$
The presence of a common ion lowers the value of the solubility product that is the common-ion effect.
If the ions are uncommon among the solute Ksp will be high due to diverse-ion effect. The presence of ion pairs will also affect Ksp value.
The reaction in the question will take place in the following way:$A{{g}_{2}}{{C}_{2}}{{O}_{4}}\rightleftharpoons 2A{{g}^{+}}+{{C}_{2}}{{O}_{4}}^{2-}$
Therefore, $2$ moles of $A{{g}^{+}}$ will give us $0.5$ moles of ${{C}_{2}}{{O}_{4}}^{2-}$. This indicates that the concentration of ${{C}_{2}}{{O}_{4}}^{2-}$ will be $0.5$ times that of $A{{g}^{+}}$ions.
The solubility product of $A{{g}_{2}}{{C}_{2}}{{O}_{4}}$ will be
${{K}_{sp}}={{[A{{g}^{+}}]}^{2}}[{{C}_{2}}O_{4}^{2-}]$
Now, we have given the concentration of $A{{g}^{+}}$ ions is $2.2\times {{10}^{-4}}mol{{L}^{-1}}$
the concentration of ${{C}_{2}}{{O}_{4}}^{2-}$ will be $0.5[A{{g}^{+}}]=0.5\times 2.2\times {{10}^{-4}}mol{{L}^{-1}}=1.1\times {{10}^{-4}}mol{{L}^{-1}}$
$\begin{align}
  & {{K}_{sp}}={{(2.2\times {{10}^{-4}}mol{{L}^{-1}})}^{2}}\times 1.1\times {{10}^{-4}}mol{{L}^{-1}} \\
 & {{K}_{sp}}=5.3\times {{10}^{-12}} \\
\end{align}$

Note: We squared the concentration of Ag ions since we obtained twice the Ag ions in the products. If it would have been three or four or any other number the power would be $3$ or $$ $4$ or the other number.