What is the common ion effect and how does it reduce solubility?

The common ion effect occurs when a solute ion already present in solution shifts the solubility equilibrium of a sparingly soluble salt. When sodium chloride is added to a lead(II) chloride solution, the increased chloride concentration causes the equilibrium to shift left, forcing more lead(II) chloride to remain undissolved. This reduces solubility because the ion product must remain constant at equilibrium.

How do you calculate molar solubility using an ICE table and Ksp?

Set up an ICE table tracking initial concentrations, changes, and equilibrium values for each ion. For lead(II) chloride in sodium chloride solution, the chloride initial concentration is 0.100 M, and the change is +2x per formula unit. Substitute equilibrium concentrations into the Ksp expression and solve for x, the molar solubility.

How does pH affect the solubility of compounds containing hydroxide ions?

Adding acid decreases pH by consuming hydroxide ions, shifting the calcium hydroxide equilibrium toward dissolution and increasing solubility. Conversely, adding base increases pH and hydroxide concentration, shifting equilibrium left and decreasing solubility. The solubility of hydroxide-containing compounds is inversely related to solution pH.

Why is selective precipitation used in wastewater treatment?

Selective precipitation removes specific contaminants like phosphate ions by adding reagents such as calcium hydroxide. The calcium ions react with phosphates to form hydroxylapatite, which precipitates out of solution. This method selectively removes target ions while leaving other dissolved salts behind, allowing filtration to separate the solid precipitate.

What is the quantitative relationship between ion concentrations at equilibrium?

The solubility product constant (Ksp) expresses the mathematical product of ion molarities at equilibrium. For lead(II) chloride, Ksp equals [Pb2+][Cl−]2. This product remains constant regardless of ion source, so increasing one ion concentration must proportionally decrease the other to maintain equilibrium.

How much does the common ion effect reduce lead(II) chloride solubility in sodium chloride solution?

In 0.100 M sodium chloride solution, lead(II) chloride solubility decreases to 1.17 × 10−3 M compared to 1.43 × 10−2 M in pure water, a 12-fold reduction. This dramatic decrease demonstrates how the common chloride ion shifts the dissolution equilibrium significantly toward the undissociated salt.

What happens to solubility when multiple calcium salts compete for calcium ions?

When calcium hydroxide is added to wastewater containing phosphates, calcium ions preferentially precipitate phosphates as hydroxylapatite because this reaction is favored. The amount of calcium added is controlled so other calcium salt solubility products are not exceeded, allowing their anions to remain dissolved and be separated later.

16.11

Factors Affecting Solubility

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The solubility of an ionic compound can vary depending on other solutes present in the solution. The solute may be an ion common to the compound, or an acid or base. Thus, two major factors affecting the solubility are the common ion effect and the pH of a solution.

If lead(II) chloride is added to a solution of sodium chloride, both salts will ionize in the water, producing sodium and lead cations and the common anion, chloride.

Because the chloride ions in solution come from both the complete ionization of sodium chloride and the partial ionization of lead(II) chloride, their concentration would be higher than that of either the sodium ions or the lead ions.

To compensate, the equilibrium between the solid lead(II) chloride and its ions in solution shifts towards the undissociated salt, causing more lead(II) chloride to remain undissolved.

Thus, the presence of a common ion decreases the solubility of a sparingly soluble substance.

For instance, the molar solubility, x, of lead(II) chloride in a 0.100 molar sodium chloride solution can be calculated from the ICE table.

The initial concentration of lead(II) ions in the solution is zero, while chloride is 0.100 molar.

Each molecule of lead(II) chloride will dissociate into one lead and two chloride ions. So, the change in concentration of lead ions will be +x and that of chloride will be +2x.

The equilibrium concentration of lead ions will be x, but for the chloride ions, it will be the sum of 2x and 0.100.

At 25 °C, the value of K_sp for lead(II) chloride is 1.17 × 10⁻⁵, and the equilibrium expression is the product of the equilibrium concentrations of lead and chloride ions, which is x(2x + 0.100)².

Because the K_sp is small, 2x is much smaller than 0.100 molar and their sum can be approximated to 0.100 molar.

Substituting into the expression yields 1.17 × 10⁻⁵= x(0.100)².

Solving for x, the molar solubility of lead(II) chloride in 0.100 molar sodium chloride solution is 1.17 × 10⁻³ molar.

Compared to the molar solubility of lead(II) chloride in water, which is 1.43 × 10⁻² molar, the common ion decreases the solubility of lead(II) chloride 12 fold.

Solubility of a compound can also be affected by the pH of the solution.

Consider the partial dissociation of calcium hydroxide into calcium ions and hydroxide. If the pH is increased, by adding potassium hydroxide for example, the common hydroxide ion will lower the solubility of calcium hydroxide.

Conversely, if the pH is decreased, for example by adding hydrochloric acid, the protons will combine with hydroxide ions, lowering the concentration of hydroxide ions in solution.

The solubility equilibrium of calcium hydroxide would then shift towards the products, leading to an increased solubility in an acidic solution.

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Compared with pure water, the solubility of an ionic compound is less in aqueous solutions containing a common ion (one also produced by dissolution o…