Question

Given,
${K_c} = 6.2 \times {10^{ - 8}}$ and ${K_{sp}}\,of\,AgCl = 1.8 \times {10^{ - 10}}\,at\,298K$
If ammonia is added to a water solution containing excess of $AgCl(s)$ only. Calculate the concentration of the complex in 1.0 $M$ aqueous ammonia.

Answer 1

Hint: The compounds which are solid in the state have some different properties at equilibrium than the ionic compounds in other states. Here ${K_c}$ is called the equilibrium constant and depends on the concentration of the reactants and products that are in equilibrium in the solution.
${K_{sp}}$ is called the solubility product which is related to the solid solute’s interaction with its saturated solution.

Complete step by step answer:
It is given that the solution is containing an excess of solid silver chloride ( $AgCl(s)$ ). $AgCl$ is in equilibrium in the solution as

The value of solubility product of $AgCl$ is given as ${K_{sp}}\,of\,AgCl = 1.8 \times {10^{ - 10}}\ ,at\,298K$
This means that for solid silver chloride in equilibrium with its saturated solution, the product of silver and chloride ions is equal to its solubility product.
In the saturated solution, the concentration of the two ions will be equal to the molar solubility of silver chloride.
Thus, assuming the molar solubility of silver chloride as $S$ we get
$[A{g^ + }] = [C{l^ - }] = S$
So the solubility product can be written as
${K_{sp}} = [A{g^ + }][C{l^ - }]$
$\Rightarrow {K_{sp}} = S \times S = {S^2}$
Equating the actual value given in the question we get

${S^2} = 1.8 \times {10^{ - 10}}$
$\Rightarrow S = 1.34 \times {10^{ - 5}}$
So we can say that $[A{g^ + }] = S = 1.34 \times {10^{ - 5}}M$
Now considering the equilibrium equation given in the question above

We are given its equilibrium constant, ${K_c} = 6.2 \times {10^{ - 8}}$
Writing the equilibrium constant equation for the above equation we get
${K_c} = \dfrac{{[A{g^ + }]{{[N{H_3}]}^2}}}{{{{[Ag{{(N{H_3})}_2}]}^ + }}}$
Therefore, upon cross multiplying we get
Concentration of the complex = $\dfrac{{[A{g^ + }]{{[N{H_3}]}^2}}}{{{K_c}}} $
$\Rightarrow$ Concentration of the complex = $\dfrac{{1.34 \times {{10}^{ - 5}} \ times {{(1)}^2}}}{{6.2 \times {{10}^{ - 8}}}}$
$\Rightarrow$ Concentration of the complex = 216.13
SSo the Concentration of the complex is 216.13 $M$

Note: Some ionic solids vary greatly when it comes to the solubility in a solution. Some salts like calcium chloride are so soluble that they absorb water from the atmosphere others like lithium fluoride are insoluble.
The solvation and saturation depend on the lattice enthalpy of the solute and the solvation enthalpy of the ions in the solution. For the salt to dissolve in the solvent the lattice energy of the salt must be overcome by the ion-ion interaction of the solute and solvent molecules.