How Does The Presence Of A Common Ion In A Solution Of A Sparingly Soluble Salt Affect The Concentration Of The Ions In The Solution, And Can You Provide An Example Of A Specific Scenario Where This Concept Would Be Particularly Relevant, Such As The Solubility Of Lead(II) Chloride In The Presence Of Sodium Chloride?

The presence of a common ion in a solution of a sparingly soluble salt significantly affects the solubility and ion concentrations due to the common ion effect, which can be explained using Le Chatelier's principle. When a common ion is added, the equilibrium shifts to minimize the disturbance, reducing the solubility of the sparingly soluble salt.

For example, consider lead(II) chloride (PbCl₂) in the presence of sodium chloride (NaCl). PbCl₂ dissociates into Pb²+ and 2Cl⁻ ions, with a solubility product (Ksp) given by:

${ K_{sp} = [Pb^{2+}][Cl^-]^2 }$

In pure water, if the solubility of PbCl₂ is ${ S }$, then:

${ [Pb^{2+}] = S }$ ${ [Cl^-] = 2S }$ ${ K_{sp} = S(2S)^2 = 4S^3 }$

When NaCl is added, the chloride ion concentration increases. For instance, if [Cl⁻] becomes 0.1 M, the new solubility ${ S' }$ is calculated as:

${ K_{sp} = S'(0.1)^2 }$ ${ S' = \frac{K_{sp}}{(0.1)^2} = \frac{1.6 \times 10^{-5}}{0.01} = 1.6 \times 10^{-3} \, \text{M} }$

Thus, [Pb²+] decreases to 0.0016 M, while [Cl⁻] increases to 0.1 M. This demonstrates that the common ion effect reduces the solubility of PbCl₂, decreasing [Pb²+] and increasing [Cl⁻].

This principle is crucial in applications like water treatment and analytical chemistry, where it is used to precipitate contaminants or ensure complete precipitation in gravimetric analysis. The effect is more pronounced when the added ion is part of the sparingly soluble salt, and the impact is amplified by the stoichiometry of the dissociation.