# M5-S8: Solubility Product

### Solubility Product Ksp

• Dissolution of ionic compounds forms an equilibrium between dissolved ions and undissolved solids.
• Dissolution of lead (II) fluoride:

$${\rm PbF}_{2(s)}\rightleftharpoons {\rm Pb}_{(aq)}^{2+}+{2F}_{(aq)}^-$$

The expression of equilibrium constant or solubility product for this reaction is:

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

 ·       The concentration of solids (ionic compounds) is not included in the expression because it stays relatively constant compared to the dissolved ions. The concentrations of dissociated ions are always expressed in mol L-1 (M).  ·       Solids and liquids are also heterogenous in that they do not disperse throughout the solution.  ·       At equilibrium, the rates of dissolution and re-formation of ionic compound are equal. This is known as the point of saturation as no more ions can dissolve in the solvent. Adding more solutes to a solution at saturation point (equilibrium) will cause precipitation to happen. Figure: Dissociation of solid PbI2 in a beaker.

• Solubility product constant varies between ionic compounds. It indicates how far the dissolution proceeds at equilibrium.
• A large solubility product value indicates high solubility as a relatively large quantity of ions are dissolved at equilibrium.
• Vice versa, a small solubility product value indicates low solubility as relatively small quantity of ions are dissolved at equilibrium.

Table: Solubility product expressions of selected few ionic compounds.

 Ionic compound Dissolution reaction Solubility product Magnesium carbonate $${\rm MgCO}_{3(s)}\rightleftharpoons{\rm Mg}_{(aq)}^{2+}+{\rm CO}_{3(aq)}^{2-}$$ $$K_{sp}=[{\rm Mg}^{2+}][{\rm CO}_3^{2-}]$$ Iron (II) hydroxide $${\rm Fe(OH)}_{2(s)}\rightleftharpoons{\rm Fe}_{(aq)}^{2+}+{\rm 2OH}_{(aq)}^-$$ $$K_{sp}=[{\rm Fe}^{2+}][{\rm OH}^-]^2$$ Calcium phosphate $${{\rm Ca}_3(PO_4)}_{2(s)}\rightleftharpoons{\rm 3Ca}_{(aq)}^{2+}+{\rm 2PO}_{4(aq)}^{3-}$$ $$K_{sp}=[{\rm Ca}^{2+}]^3[{\rm PO}_4^{3-}]^2$$

Example 1

The following table shows the solubility product constants of selected ionic compounds at 25ºC.

 Name, formula Ksp Aluminium hydroxide, Al(OH)3 3 xx 10^(-34) Lead(II) fluoride, PbF2 3.6 xx 10^(-8) Silver sulfide, Ag2S 8.0 xx 10^(-48) Cobalt(II) carbonate, CoCO3 1.0 xx 10^(-10)

Rank the ionic compounds in the table in order of increasing solubility.

### Solubility of Ionic Compounds

• The solubility of an ionic compound is usually expressed as the amount of solute dissolved per volume of solvent. This can be either expressed as mass per volume e.g. g/100 mL or as molarity, mol L-1 (M) .
• When solubility of an ionic compound is known, its Ksp can be determined. For example, at 25ºC, the solubility of PbF2 is found to be 0.64 g/L. Calculate the Ksp of PbF2.

 Step 1: Convert solubility into mol L-1 by dividing by the molar mass of the ionic compound. In this case, the molar mass of PbF2 = 245.2 g mol-1 $$n=\frac{0.64\ g}{245.2\ g\ {\rm mol}^{-1}}$$ $$n=0.0026\ mol$$ Therefore, solubility = 0.0026 mol L-1 (M) Step 2: Write an equation representing the dissociation of the ionic compound. $${\rm PbF}_{2(s)}\rightleftharpoons{\rm Pb}_{(aq)}^{2+}+{\rm 2F}_{(aq)}^-$$ Step 3: Write an expression for the solubility product of the ionic compound and find the concentration of each ion. $$K_{sp}=[{\rm Pb}^{2+}][{\rm F}^-]^2$$  $$K_{sp}=[0.0026][2×0.0026]^2$$ $$K_{sp}=7.1\times{10}^{-8}$$

• Using Ksp values from the data sheet, solubility of various ionic compounds can be determined.
 Step 1: Write an equation representing the dissociation of the ionic compound. $${\rm PbI}_{2(s)}\rightleftharpoons{\rm Pb}_{(aq)}^{2+}+{\rm 2I}_{(aq)}^-$$ Step 2: Write an expression for the solubility product of the ionic compound. $$K_{sp}=[{\rm Pb}^{2+}][{\rm I}^-]^2$$ Step 3: Let s be the solubility of the ionic compound. Assign concentration of ions in terms of s. $${\rm PbI}_{2(s)}\rightleftharpoons{\rm Pb}_{(aq)}^{2+}+{\rm 2I}_{(aq)}^-$$             s              s            2s Step 4: Substitute Ksp (from data sheet) and concentrations of ions in terms of s into Ksp expression. $$9.8\times{10}^{-9}=[s][2s]^2$$ Step 5: Solve for solubility (s). $$9.8\times{10}^{-9}=4s^3$$ $$s=1.3\times{10}^{-3} mol L^{-1}$$

## Practice Questions

Determining Ksp from solubility

Question 1

(a) Lead(II) sulfate (PbSO4) is a key component in lead acid car batteries.

(b) Its solubility in water at 25°C is 4.25 xx 10-3 g/100 mL. What is the Ksp of PbSO4?

(c) In terms of their constituent ions, explain why lead(II) fluoride has a much greater solubility than lead(II) sulfate.

Question 2

The solubility of silver chloride in water at 25ºC is 1.34 xx 10-5.What is the Ksp of silver chloride?

Question 3

The solubility of calcium hydroxide in water at 25ºC is 0.074 g/100 mL. What is the Ksp of calcium hydroxide?

Determining solubility from Ksp

Question 1

Ksp of barium sulfate is 11.1 xx 10-10. What is the solubility of barium sulfate in moles per litre? What about in grams per mL?

Question 2

Ksp of iron(III) hydroxide is 14.4 xx 10-38. What is the solubility of iron(III) hydroxide in moles per litre? What about in grams per mL?

### Quotient Solubility Product Q

• Similar to equilibrium quotient Q, the solubility quotient is the value assigned to a dissociation reaction when it is not at equilibrium.
• When Q < Ksp, the dissociation system has not reached the point of saturation, it is unsaturated. This means more ions will dissolve and become hydrated by solvent molecules e.g. water.
• When Q > Ksp, the dissociation system has exceeded the point of saturation, it is saturated. This means precipitate will form as no more ions can be hydrated by solvent molecules.
• Important: high Ksp indicates more ions can dissolve before saturation (equilibrium) is reached. This means high solubility. In contrast, low Ksp indicates less ions can dissolve before saturation is reached, low solubility.

## Practice Questions

Determining formation of precipitate

Question 1

0.100 g of BaSO4 is added to 500.0 mL of water at 25 ºC. Calculate the Qsp of this barium sulfate solution and thus determine whether a precipitate will form.

Question 2

Will lead(II) chloride precipitate when 50 mL of 0.10 M Pb(NO3)2 solution is mixed with 50 mL of 0.10 M NaCl solution? Support your answer with a balanced chemical equation and calculations.

Question 3

Equal volumes of 0.25 mol L–1 solutions of silver nitrate and calcium chloride are mixed at 25 ºC. Predict whether a precipitate will form. Support your answer with calculations.

### Common Ion Effect

• According to Le Châtelier’s principle, changing the concentration of chemical species of an equilibrium system will disturb and move the position of the equilibrium. Addition of an ionic compound that contains an ion present in the equilibrium system will achieve the same result. This is known as the common ion effect.
• For example, consider a saturated solution of lead(II) chromate:

$$PbCrO_{4(s)}\rightleftharpoons {\rm Pb}_{(aq)}^{2+}+CrO_{4(aq)}^{2-}\ \ \ \ K_{sp}=\left[{\rm Pb}_{\left(aq\right)}^{2+}\right]\left[CrO_{4\left(aq\right)}^{2-}\right]=2.3\times{10}^{-13}$$

At a given temperature, Ksp depends on the product of the ion concentrations. If the concentration of either ion goes up, the other goes down to maintain the value of Ksp.

Suppose we add Na2CrO4, a soluble salt, to the saturated PbCrO4 solution. The concentration of the common ion, CrO42-, increases, and some of it combines with Pb2+ ion to form more solid PbCrO4 (figure below). In effect, the addition of Na2CrO4 will shift the equilibrium to the left. As a result of the addition, [Pb2+] is lower. And, since [Pb2+] defines the solubility of PbCrO4, in effect, the solubility of PbCrO4 has decreased. Note that we would get the same result if Na2CrO4 solution were the solvent; that is, PbCrO4 is more soluble in water than in aqueous Na2CrO4.

• Addition of a common ion reduces the solubility of the ionic compound.
• Removal of a common ion increases the solubility of the ionic compound.
• Adding or removal common ions does not change the value of the solubility product, Ksp.

## Practice Questions

Question 1

Given the Ksp for calcium sulfate is 9.1 xx 10-6, determine whether a solution which contains calcium ions at a concentration of 1.00 mol/L and sulfate ions at a concentration of 2.56 mol/L is saturated with calcium sulfate.

Question 2

Given that Ksp BaSO4 = [Ba2+] [SO42-] = 1.1 xx 10-10, determine the solubility of barium sulfate in a solution of sodium sulfate of a concentration 1.00 mol/L.

### Effect of Temperature on Solubility

• Since the equilibrium constant changes with constant, the value of solubility product Ksp also changes with temperature.
• If dissolution of an ionic compound is endothermic, an increase in temperature will increase Ksp which in turn increases the solubility of the solution. The effect is opposite if temperature is reduced.
• If dissolution of an ionic compound is exothermic, a decrease in temperature will increase Ksp which in turn increases the solubility of the solution. The effect is opposite if temperature is increased.

Table: Effect of changing temperature on the solubility of ionic compounds.

 Endothermic dissolution Exothermic dissolution Increase in temperature Ksp increases, solubility increases Ksp decreases, solubility decreases Decrease in temperature Ksp decreases, solubility decreases Ksp increases, solubility increases

## Practice Question

Consider the Ksp values of AgCl and BaSO4 at 25 ºC.

The Ksp of AgCl at 100ºC is 2.15 ´ 10–8.

The Ksp of BaSO4 at 18ºC is 1 ´ 10–10.

Explain whether the dissolution of AgCl and BaSO4 is endothermic or exothermic.

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