E. Chemical Systems and Equilibrium · Acid-Base and Solubility EquilibriaSolubility Equilibria and the Solubility Product, Ksp
E3.4 E2.4 E2.1About 70 minutes
Learning goals- I can describe a saturated solution of a slightly soluble ionic compound as a dynamic equilibrium.
- I can write Ksp expressions for slightly soluble ionic compounds.
- I can calculate molar solubility from Ksp, and Ksp from solubility data.
- I can explain when Ksp values can and cannot be used to compare solubilities directly.
Words to know
solubility product constant, Ksp
The equilibrium constant for a slightly soluble ionic solid dissolving into its ions, for example Ksp = [Ag⁺][Cl⁻] for AgCl.
molar solubility (s)
The number of moles of a solute that dissolve to make one litre of saturated solution, in mol/L.
solubility
The amount of solute that dissolves in a given amount of solvent at a given temperature, often in g/L or g/100 mL.
saturated solution
A solution in dynamic equilibrium with undissolved solute: the rates of dissolving and precipitating are equal.
"Insoluble" is never quite zero
In Grade 11 you used a solubility table to call silver chloride "insoluble" (or of low solubility) and predicted that it precipitates. But if you shake solid AgCl in water, a tiny amount does dissolve: about 2 milligrams per litre. That small amount matters. It decides whether a toxic metal stays in a lake, whether kidney stones form, how fluoride protects teeth, and how much barium a patient absorbs during a medical X-ray. This lesson gives "slightly soluble" a number, the solubility product constant, Ksp.
Remember from Grade 11
In Solutions and Solubility (SCH3U) you used a solubility table to predict precipitates, wrote net ionic equations such as Ag⁺(aq) + Cl⁻(aq) → AgCl(s), and defined a saturated solution. From Chemical Equilibrium this year: solids and pure liquids are left out of equilibrium expressions.
A saturated solution is a dynamic equilibrium
Put excess AgCl in water. Ions leave the crystal surface (dissolving) and ions in solution collide with the surface and rejoin it (precipitating). At first, dissolving is faster. As ions build up, precipitation speeds up, until the two rates are equal. The solution is now saturated, and the amounts of solid and dissolved ions stay constant even though both processes continue.
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) Ksp = [Ag⁺][Cl⁻]
The solid is left out (its concentration doesn't change), so Ksp is just the product of the ion concentrations, each raised to its coefficient. For example:
- CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq) Ksp = [Ca²⁺][F⁻]²
- Ag₂CrO₄(s) ⇌ 2 Ag⁺(aq) + CrO₄²⁻(aq) Ksp = [Ag⁺]²[CrO₄²⁻]
- Ca₃(PO₄)₂(s) ⇌ 3 Ca²⁺(aq) + 2 PO₄³⁻(aq) Ksp = [Ca²⁺]³[PO₄³⁻]²
Ksp is used only for slightly soluble compounds. For very soluble salts such as NaCl, the solutions are so concentrated that ions interact strongly and a simple Ksp does not describe them well.
| Compound | Ksp at 25 °C | Compound | Ksp at 25 °C |
|---|
| AgCl | 1.8 × 10⁻¹⁰ | CaF₂ | 3.5 × 10⁻¹¹ |
| AgBr | 5.4 × 10⁻¹³ | Ca(OH)₂ | 5.0 × 10⁻⁶ |
| AgI | 8.5 × 10⁻¹⁷ | CaSO₄ | 4.9 × 10⁻⁵ |
| Ag₂CrO₄ | 1.1 × 10⁻¹² | CaC₂O₄ (calcium oxalate) | 2.3 × 10⁻⁹ |
| BaSO₄ | 1.1 × 10⁻¹⁰ | Ca₃(PO₄)₂ | 2.1 × 10⁻³³ |
| BaCO₃ | 2.6 × 10⁻⁹ | Mg(OH)₂ | 5.6 × 10⁻¹² |
| CaCO₃ | 3.4 × 10⁻⁹ | Fe(OH)₃ | 2.8 × 10⁻³⁹ |
| PbCl₂ | 1.7 × 10⁻⁵ | PbI₂ | 9.8 × 10⁻⁹ |
| PbSO₄ | 2.5 × 10⁻⁸ | Cu(OH)₂ | 2.2 × 10⁻²⁰ |
Molar solubility and Ksp
Let s be the molar solubility: the moles of solid that dissolve per litre. The ion concentrations follow from the formula. For CaF₂, each formula unit that dissolves gives one Ca²⁺ and two F⁻, so [Ca²⁺] = s and [F⁻] = 2s, and Ksp = s(2s)² = 4s³.
Worked exampleExample 1: Molar solubility from Ksp (AB type)
Find the molar solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰) and its solubility in g/L. M(AgCl) = 143.32 g/mol.
- AgCl(s) ⇌ Ag⁺ + Cl⁻; [Ag⁺] = [Cl⁻] = s.
- Ksp = s² = 1.8 × 10⁻¹⁰, so s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ mol/L.
- Solubility = 1.34 × 10⁻⁵ mol/L × 143.32 g/mol = 1.9 × 10⁻³ g/L.
s = 1.3 × 10⁻⁵ mol/L, about 1.9 mg per litre.
Worked exampleExample 2: Molar solubility from Ksp (AB₂ type)
Find the molar solubility of CaF₂ (Ksp = 3.5 × 10⁻¹¹) and [F⁻] in a saturated solution.
- CaF₂(s) ⇌ Ca²⁺ + 2 F⁻; [Ca²⁺] = s, [F⁻] = 2s.
- Ksp = s(2s)² = 4s³ = 3.5 × 10⁻¹¹, so s³ = 8.75 × 10⁻¹² and s = ∛(8.75 × 10⁻¹²) = 2.1 × 10⁻⁴ mol/L.
- [F⁻] = 2s = 4.1 × 10⁻⁴ mol/L.
s = 2.1 × 10⁻⁴ mol/L, [F⁻] = 4.1 × 10⁻⁴ mol/L. (Use the x^(1/3) or ∛ key; a common mistake is forgetting to divide by 4.)
Worked exampleExample 3: Ksp from a measured solubility
The solubility of PbI₂ is 0.62 g/L at 25 °C. M(PbI₂) = 461.0 g/mol. Find Ksp.
- s = 0.62 g/L ÷ 461.0 g/mol = 1.34 × 10⁻³ mol/L.
- PbI₂(s) ⇌ Pb²⁺ + 2 I⁻; [Pb²⁺] = 1.34 × 10⁻³, [I⁻] = 2.69 × 10⁻³ mol/L.
- Ksp = [Pb²⁺][I⁻]² = (1.34 × 10⁻³)(2.69 × 10⁻³)² = 9.7 × 10⁻⁹.
Ksp = 9.7 × 10⁻⁹, which agrees with the table value (9.8 × 10⁻⁹) to within rounding.
Worked exampleExample 4: The pH of milk of magnesia
Milk of magnesia is a suspension of Mg(OH)₂ in water. Find the pH of the saturated solution around the solid. Ksp = 5.6 × 10⁻¹².
- Mg(OH)₂(s) ⇌ Mg²⁺ + 2 OH⁻; Ksp = 4s³ = 5.6 × 10⁻¹², so s = ∛(1.4 × 10⁻¹²) = 1.12 × 10⁻⁴ mol/L.
- [OH⁻] = 2s = 2.24 × 10⁻⁴ mol/L; pOH = 3.65.
- pH = 14.00 − 3.65 = 10.35.
pH ≈ 10.35: basic, but mild enough to use as an antacid because so little dissolves at once.
Comparing solubilities: be careful
For compounds of the same formula type (all AB, or all AB₂), a smaller Ksp means a smaller molar solubility. AgI (8.5 × 10⁻¹⁷) is less soluble than AgBr (5.4 × 10⁻¹³), which is less soluble than AgCl (1.8 × 10⁻¹⁰).
For different types, compare molar solubilities, not Ksp values. Ag₂CrO₄ has a smaller Ksp (1.1 × 10⁻¹²) than AgCl (1.8 × 10⁻¹⁰), but its molar solubility, s = ∛(Ksp ÷ 4) = 6.5 × 10⁻⁵ mol/L, is about five times larger than that of AgCl (1.3 × 10⁻⁵ mol/L).
Limits: calculated solubilities are approximate. Some ions pair up in solution or react with water (CO₃²⁻, PO₄³⁻, S²⁻ and Fe³⁺ all do), which changes the real solubility. Like any K, Ksp changes with temperature; most solids become more soluble when heated, but a few, such as Ca(OH)₂, become less soluble.
Just for funSilver chloride is so insoluble that a litre of water dissolves only about 1.9 mg of it, about a tenth of the mass of a single grain of rice. A chemist's "soluble" and a cook's "soluble" are very different things.
Hands-on activityGrow crystals from a saturated solution
You need: Epsom salt (magnesium sulfate, from a pharmacy) or table salt, a heat-safe jar, hot tap water (or kettle water poured by an adult), a spoon, a pencil, a piece of cotton string, a kitchen scale if you have one, and a dark corner where the jar can sit undisturbed for several days.
Grown-up note: an adult should pour any very hot water. Epsom salt is a pharmacy product; don't eat it, and wash your hands after handling.
- Fill the jar about half full with hot water. Stir in Epsom salt a spoonful at a time until some stays undissolved at the bottom even after a minute of stirring. Record roughly how much dissolved.
- Tie the string to the pencil and hang it in the jar so it does not touch the bottom.
- Leave the jar for 2 to 5 days. Observe the crystals on the string and the solid at the bottom each day.
- Explain your observations using the words saturated, dynamic equilibrium, dissolving and precipitating.
What to look for: as the water cools, solubility falls, so the solution becomes supersaturated and crystals grow on the string and at the bottom. Once the jar reaches room temperature and stays saturated, the crystals keep changing shape slowly (small ones can shrink while big ones grow) even though the total amount of solid stays about the same: ions are still dissolving and precipitating at equal rates. (Epsom salt is quite soluble, so we don't use Ksp for it; the idea of dynamic equilibrium is the same as for slightly soluble salts.)
Your turn
1Write the dissolving equilibrium and the Ksp expression for BaSO₄.
BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = [Ba²⁺][SO₄²⁻].
2Write the Ksp expression for Fe(OH)₃.
Fe(OH)₃(s) ⇌ Fe³⁺(aq) + 3 OH⁻(aq); Ksp = [Fe³⁺][OH⁻]³.
3Find the molar solubility of BaSO₄ (Ksp = 1.1 × 10⁻¹⁰).
s² = 1.1 × 10⁻¹⁰, s = 1.0 × 10⁻⁵ mol/L.
4Find the molar solubility of Ag₂CrO₄ (Ksp = 1.1 × 10⁻¹²) and [Ag⁺] in the saturated solution.
Ksp = (2s)²s = 4s³; s = ∛(2.75 × 10⁻¹³) = 6.5 × 10⁻⁵ mol/L; [Ag⁺] = 2s = 1.3 × 10⁻⁴ mol/L.
5The molar solubility of CaSO₄ is 7.0 × 10⁻³ mol/L. Find Ksp.
Ksp = s² = (7.0 × 10⁻³)² = 4.9 × 10⁻⁵.
6Which is more soluble in mol/L: AgCl (Ksp 1.8 × 10⁻¹⁰) or AgBr (Ksp 5.4 × 10⁻¹³)? Explain.
AgCl. Both are the AB type, so the larger Ksp means the larger molar solubility.
More practice: warm-up
1Why is the solid left out of a Ksp expression?
Its concentration (amount per volume of the solid itself) doesn't change, so it is part of the constant.
2In a saturated solution of CaF₂ with molar solubility s, what is [F⁻]?
3Write Ksp in terms of s for an AB₂ compound.
4A larger Ksp (for compounds of the same type) means a ___ solubility.
5Name one compound whose solubility decreases as temperature rises.
Ca(OH)₂ (calcium hydroxide).
More practice: on your own
1Find the solubility of CaCO₃ in g/L (Ksp = 3.4 × 10⁻⁹, M = 100.09 g/mol).
s = √(3.4 × 10⁻⁹) = 5.8 × 10⁻⁵ mol/L; × 100.09 = 5.8 × 10⁻³ g/L (about 6 mg/L).
2Find the pH of a saturated Ca(OH)₂ solution (Ksp = 5.0 × 10⁻⁶).
4s³ = 5.0 × 10⁻⁶, s = 1.08 × 10⁻² mol/L; [OH⁻] = 2.15 × 10⁻² mol/L; pOH = 1.67, pH ≈ 12.33.
3The solubility of PbCl₂ is 4.5 g/L (M = 278.1 g/mol). Find Ksp.
s = 4.5 ÷ 278.1 = 1.62 × 10⁻² mol/L. Ksp = 4s³ = 4(1.62 × 10⁻²)³ = 1.7 × 10⁻⁵.
4Find the molar solubility of Fe(OH)₃ (Ksp = 2.8 × 10⁻³⁹), ignoring the OH⁻ from water. Then explain why this answer is not realistic.
27s⁴ = 2.8 × 10⁻³⁹, s⁴ = 1.04 × 10⁻⁴⁰, s = 1.0 × 10⁻¹⁰ mol/L. This would give [OH⁻] = 3.0 × 10⁻¹⁰ mol/L, far less than the 1.0 × 10⁻⁷ mol/L already present in neutral water, so the water's OH⁻ can't be ignored (and Fe³⁺ also reacts with water). The real solubility is even lower.
5Ca₃(PO₄)₂ has Ksp = 2.1 × 10⁻³³ and CaF₂ has Ksp = 3.5 × 10⁻¹¹. Can you say which is more soluble by comparing Ksp values? Calculate both molar solubilities.
No, they are different formula types. CaF₂: s = ∛(3.5 × 10⁻¹¹ ÷ 4) = 2.1 × 10⁻⁴ mol/L. Ca₃(PO₄)₂: Ksp = (3s)³(2s)² = 108s⁵, s = ⁵√(2.1 × 10⁻³³ ÷ 108) = 1.1 × 10⁻⁷ mol/L. CaF₂ is more soluble.
Challenge
1Derive the Ksp expression in terms of s for Ca₃(PO₄)₂ and show that Ksp = 108s⁵.
Each formula unit gives 3 Ca²⁺ and 2 PO₄³⁻: [Ca²⁺] = 3s, [PO₄³⁻] = 2s. Ksp = (3s)³(2s)² = 27s³ × 4s² = 108s⁵.
2A saturated solution of a metal hydroxide M(OH)₂ has pH 9.70. Find Ksp.
pOH = 4.30, [OH⁻] = 5.01 × 10⁻⁵ mol/L. [M²⁺] = ½[OH⁻] = 2.51 × 10⁻⁵ mol/L. Ksp = (2.51 × 10⁻⁵)(5.01 × 10⁻⁵)² = 6.3 × 10⁻¹⁴.
3A patient drinks a suspension of BaSO₄ before an X-ray. Ba²⁺ ions are toxic. Use Ksp (1.1 × 10⁻¹⁰) to explain why the procedure is considered safe, and calculate [Ba²⁺] in a saturated solution.
[Ba²⁺] = s = √(1.1 × 10⁻¹⁰) = 1.0 × 10⁻⁵ mol/L, which is only about 1.4 mg of Ba²⁺ per litre. Because BaSO₄ is so slightly soluble, very little toxic Ba²⁺ dissolves, while the dense solid still blocks X-rays and outlines the digestive tract. (Soluble barium compounds, such as BaCl₂, would be dangerous.)
Quiz
1. What is the Ksp expression for Ag₂CO₃?
2. For an AB₂ compound, Ksp in terms of molar solubility s is
3. The molar solubility of an AB compound is 2.0 × 10⁻⁴ mol/L. What is Ksp?
4. Which statement about a saturated solution in contact with excess solid is true?
5. Ag₂CrO₄ (Ksp 1.1 × 10⁻¹²) and AgCl (Ksp 1.8 × 10⁻¹⁰): which has the greater molar solubility?
6. What is the molar solubility of CaF₂ (Ksp = 3.5 × 10⁻¹¹)?
Big idea (E3.4, E2.4): a saturated solution of a slightly soluble ionic compound is a dynamic equilibrium. Its constant, Ksp, is the product of the ion concentrations raised to their coefficients. Molar solubility s and Ksp can be calculated from each other, using the formula to relate each ion's concentration to s.
How to help: For each compound, ask your teen to write the dissolving equation first and say, "For every one that dissolves, I get ___ of these and ___ of those." That sentence prevents most errors. The crystal activity needs an adult for hot water; check on the jar together each day and ask what is happening at the particle level.
Common difficulties:
- Forgetting to raise the doubled concentration to a power: for CaF₂, Ksp = s(2s)², not s(2s).
- Comparing Ksp values directly for compounds with different formula types.
- Taking a square root when a cube root (or fourth root) is needed.
- Forgetting to convert between g/L and mol/L.