Because the solubility of an ionic compound depends on the product of the concentrations of the ions, this solubility can be greatly affected if there are already some of those ions present in the solution. For example, imagine we have a 0.1 molar solution of sodium chloride. This solution has a [Na+] = [Cl-1] = 0.1 M. Now we try to dissolve some AgCl into this same solution. Silver chloride is fairly insoluble. It has a solubility in pure water of about 2 mg/L. That corresponds to a molar solubility of 1.3 x 10-5 moles/L. Given as a Ksp this is 1.8 x 10-10. However, now the amount of silver chloride that will dissolve will be less. This is because there is a common ion in the solution, chloride. The chloride concentration is already 0.1 M. Therefore the silver concentration will be even less than it would be in the solution of pure water. The silver ion concentration will be reflective of the amount of silver chloride that dissolves since this is the only source of silver ions.
We can find the concentration of silver ions at equilibrium from Ksp since we already know the chloride ion concentration. We know that only a tiny amount of silver chloride will dissolve. This means the chloride concentration will stay around 0.1 M. We can then use Ksp to solve for the silver ion concentration.
\[K_{sp} = \rm{[Ag^+][Cl^-] = [Ag^+] = }K_{sp}/\rm{[Cl^-]} = (1.8 x 10^{-10})/(0.1) = 1.8 x 10^{-9}\]
Therefore, in the 0.1 M solution of NaCl only 1.8 x 10-9 moles of AgCl dissolve in one L. This is approximately 10,000 times less than would dissolve in pure water (1.3 x 10-5 M). We would reduce this amount even further by adding more chloride ions to the solution by adding more of the extremely soluble NaCl. In this way, it is possible to manipulate the solubility of different compounds by changing the concentrations of common ions. This can be done to not only change the solubility, but to form precipitates.
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