ChemistryAS 91392NCEA Level 3 Science

Solubility and Ks

What this lesson covers

A saturated solution of a sparingly soluble ionic solid is a dynamic equilibrium, described by the solubility product Ks.
Writing Ks expressions and converting between Ks and solubility s for AB, A₂B and AB₂ solids.
Why Ks values can be compared directly only for solids of the same type.
The common ion effect, and how complex ion formation or acid reacting with a basic anion increases solubility.
Predicting precipitation by comparing the ionic product Q with Ks, after allowing for dilution.

Key idea

A saturated solution is an equilibrium

A sparingly soluble solid such as silver chloride dissolves only a tiny amount. Once the solution is saturated, ions leave the solid and rejoin it at the same rate: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).
The equilibrium constant for this is the solubility product, Ks = [Ag⁺][Cl⁻]. The solid is left out of the expression, as solids always are.
Ks depends only on temperature. A small Ks means very few ions in solution at equilibrium.
Solubility, s, is the amount of solid that dissolves to make a saturated solution, in mol L⁻¹. It is not the same number as Ks, but each can be found from the other.

Key idea

Ks in terms of s for three types of solid

If s mol of solid dissolves per litre, the ion concentrations follow from the formula: AB gives s of each ion; A₂B gives 2s of A and s of B; AB₂ gives s of A and 2s of B.
Substitute into Ks, remembering to square the concentration of the ion with the 2 in front: (2s)² = 4s².
These relationships assume the ions do not react further with water, which is how Level 3 calculations treat them.

Type	Example	Equation	Ks in terms of s
AB	AgCl	AgCl ⇌ Ag⁺ + Cl⁻	Ks = s × s = s²
A₂B	Ag₂CrO₄	Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻	Ks = (2s)² × s = 4s³
AB₂	PbI₂	PbI₂ ⇌ Pb²⁺ + 2I⁻	Ks = s × (2s)² = 4s³

Worked example

Example: solubility from Ks, and Ks from solubility

At 25 °C, Ks(AgCl) = 1.8 × 10⁻¹⁰ and Ks(Ag₂CrO₄) = 1.1 × 10⁻¹². Calculate the solubility of each, and say which dissolves more.
The solubility of PbI₂ at 25 °C is 1.35 × 10⁻³ mol L⁻¹. Calculate Ks(PbI₂).

AgCl: s² = 1.8 × 10⁻¹⁰, so s = √(1.8 × 10⁻¹⁰)
s(AgCl) = 1.3 × 10⁻⁵ mol L⁻¹
Ag₂CrO₄: 4s³ = 1.1 × 10⁻¹², so s³ = 2.75 × 10⁻¹³[Ag⁺] = 2s and [CrO₄²⁻] = s
s(Ag₂CrO₄) = ∛(2.75 × 10⁻¹³) = 6.5 × 10⁻⁵ mol L⁻¹
Ag₂CrO₄ has the smaller Ks but is about five times more solubledifferent types, so compare s, not Ks
PbI₂: [Pb²⁺] = s, [I⁻] = 2s, so Ks = 4s³ = 4 × (1.35 × 10⁻³)³
Ks(PbI₂) = 9.84 × 10⁻⁹

Key idea

Calculator: try the cube roots

Key idea

The common ion effect

If the water already contains one of the solid's ions, the solid dissolves less. Adding Cl⁻ to AgCl(s) ⇌ Ag⁺ + Cl⁻ shifts the equilibrium to the left, by Le Chatelier's principle, so less AgCl can dissolve.
In the calculation, the added ion's concentration is used for that ion, because the tiny amount from the solid adds almost nothing to it.
Ks does not change, because the temperature has not changed; only the solubility does.

Worked example

Example: solubility with a common ion

Calculate the solubility of AgCl in 0.010 mol L⁻¹ NaCl, and of Mg(OH)₂ in a solution of pH 12.00. Ks(AgCl) = 1.8 × 10⁻¹⁰ and Ks(Mg(OH)₂) = 5.6 × 10⁻¹².

AgCl: [Cl⁻] = 0.010 mol L⁻¹, so s = [Ag⁺] = Ks ÷ [Cl⁻] = 1.8 × 10⁻¹⁰ ÷ 0.010Cl⁻ comes almost all from the NaCl
s = 1.8 × 10⁻⁸ mol L⁻¹, about 750 times less than in water
Mg(OH)₂: pH 12.00, so pOH = 2.00 and [OH⁻] = 1.0 × 10⁻² mol L⁻¹pH + pOH = 14
s = [Mg²⁺] = Ks ÷ [OH⁻]² = 5.6 × 10⁻¹² ÷ (1.0 × 10⁻²)²
s = 5.6 × 10⁻⁸ mol L⁻¹

Key idea

Making a solid dissolve more

Anything that removes one of the ions from solution shifts the equilibrium to the right, so more solid dissolves.
Complex ions: AgCl dissolves in concentrated ammonia because Ag⁺ + 2NH₃ ⇌ [Ag(NH₃)₂]⁺ removes Ag⁺. Zinc hydroxide dissolves in excess NaOH because Zn²⁺ + 4OH⁻ ⇌ [Zn(OH)₄]²⁻ removes Zn²⁺.
Acids: a basic anion such as OH⁻ or CO₃²⁻ reacts with H₃O⁺, so hydroxides and carbonates dissolve in acid. Mg(OH)₂ in antacid tablets dissolves in stomach acid as OH⁻ + H₃O⁺ → 2H₂O.
At Waitomo, rainwater containing dissolved CO₂ is slightly acidic, so it dissolves limestone (CaCO₃) as the CO₃²⁻ is turned into HCO₃⁻. Inside the cave CO₂ escapes from the drips, the reaction reverses, and CaCO₃ precipitates as stalactites.

Key idea

Try it: more or less soluble?

Key idea

Will a precipitate form? Compare Q with Ks

The ionic product, Q, uses the same expression as Ks but with the concentrations just after mixing, before anything precipitates.
Mixing two solutions dilutes both, so first find each ion's new concentration: c(new) = c(old) × V(old) ÷ V(total).
If Q > Ks, the solution holds more ions than it can at equilibrium, so a precipitate forms until Q falls to Ks. If Q ≤ Ks, no precipitate forms.
Show the comparison in your answer: write the value of Q, the value of Ks, and which is larger.

A scale of the ionic product Q with Ks marked in the middle: to the left Q is less than Ks and no precipitate forms; at Ks the solution is just saturated; to the right Q is greater than Ks and a precipitate forms

Worked example

Example: mixing magnesium chloride and sodium hydroxide

50.0 mL of 0.010 mol L⁻¹ MgCl₂ is mixed with 50.0 mL of NaOH solution of pH 10.00. Will Mg(OH)₂ precipitate? Ks(Mg(OH)₂) = 5.6 × 10⁻¹².

NaOH: pOH = 4.00, so [OH⁻] = 1.0 × 10⁻⁴ mol L⁻¹ before mixing
total volume 100.0 mL, so each concentration halvesc × 50.0 ÷ 100.0
[Mg²⁺] = 5.0 × 10⁻³ and [OH⁻] = 5.0 × 10⁻⁵ mol L⁻¹
Q = [Mg²⁺][OH⁻]² = 5.0 × 10⁻³ × (5.0 × 10⁻⁵)² = 1.3 × 10⁻¹¹
Q = 1.3 × 10⁻¹¹ > Ks = 5.6 × 10⁻¹², so a precipitate forms
forgetting to halve the concentrations would give Q eight times too big

Key idea

What Achieved, Merit and Excellence look like

Achieved writes the equation and Ks expression and states the change. Merit explains the change with the equilibrium shift. Excellence links every species to the shift, with equations, and says what happens to s and to Ks.

Question: explain the effect of adding dilute HCl to a saturated solution of Mg(OH)₂ in contact with solid.
Grade	A model answer
Achieved	Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq). The acid makes more Mg(OH)₂ dissolve.
Merit	The H₃O⁺ from the acid reacts with OH⁻: H₃O⁺ + OH⁻ → 2H₂O. This lowers [OH⁻], so the equilibrium shifts to the right to replace it, and more Mg(OH)₂ dissolves.
Excellence	Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq). Added H₃O⁺ neutralises OH⁻ (H₃O⁺ + OH⁻ → 2H₂O), lowering [OH⁻] so that Q = [Mg²⁺][OH⁻]² falls below Ks. The equilibrium shifts right to restore Q = Ks, so more solid dissolves and [Mg²⁺] rises: the solubility increases. Ks is unchanged because the temperature is constant.

Common mistakes

For PbI₂, Ks = s × 2s = 2s².
The iodide concentration is 2s and it is squared: Ks = s × (2s)² = 4s³.
Ag₂CrO₄ has a smaller Ks than AgCl, so it must be less soluble.
Ks values compare solubility only for solids of the same type; work out s for each.
Adding NaCl to saturated AgCl lowers Ks.
Only temperature changes Ks; the common ion lowers the solubility, s.
Q is worked out from the concentrations of the two solutions before they are mixed.
Mixing dilutes both, so find each ion's concentration in the total volume first.

Try it at home

Dissolve a sparingly soluble solid with acid

You need: 2 clean, dry eggshells, white vinegar, tap water, 2 clear glasses, a spoon

Rinse the eggshells, peel off the thin inner skin, and break each into a few pieces.
Put one shell's pieces in a glass of water and the other's in a glass of vinegar, each just covered.
Look closely at both glasses for 5 minutes, then check again after a few hours and the next day.
Look for: in water nothing seems to happen, because calcium carbonate is only sparingly soluble. In vinegar bubbles of CO₂ form on the shell and it slowly dissolves, because H₃O⁺ from the acid reacts with the CO₃²⁻ ions, so CaCO₃(s) ⇌ Ca²⁺ + CO₃²⁻ keeps shifting to the right.

Safety: Raw eggshell can carry salmonella, so wash your hands after handling it, and do not drink either liquid.

Your turn

Write the Ks expression for PbCl₂.

Ks(BaSO₄) = 1.1 × 10⁻¹⁰ at 25 °C. Calculate its solubility.

The solubility of CaF₂ is 2.1 × 10⁻⁴ mol L⁻¹. Calculate Ks(CaF₂).

What happens to the solubility of AgCl when concentrated ammonia is added, and why?

Equal volumes of 2.0 × 10⁻⁴ mol L⁻¹ AgNO₃ and 2.0 × 10⁻⁴ mol L⁻¹ NaCl are mixed. Does AgCl precipitate? Ks(AgCl) = 1.8 × 10⁻¹⁰.

Show answer

1. Ks = [Pb²⁺][Cl⁻]². 2. s = √(1.1 × 10⁻¹⁰) = 1.0 × 10⁻⁵ mol L⁻¹. 3. Ks = 4s³ = 4 × (2.1 × 10⁻⁴)³ = 3.7 × 10⁻¹¹. 4. It increases: Ag⁺ + 2NH₃ → [Ag(NH₃)₂]⁺ removes Ag⁺, so the equilibrium shifts right and more AgCl dissolves. 5. After mixing each ion is 1.0 × 10⁻⁴ mol L⁻¹, so Q = 1.0 × 10⁻⁸, which is greater than Ks = 1.8 × 10⁻¹⁰: yes, a precipitate forms.

Your turn

Exam-style question: zinc hydroxide

Zinc hydroxide, Zn(OH)₂, is a sparingly soluble white solid. Ks(Zn(OH)₂) = 3.0 × 10⁻¹⁷ at 25 °C.

Write the equation for the equilibrium in a saturated solution, and the Ks expression.

Calculate the solubility of Zn(OH)₂ in pure water.

Calculate its solubility in a solution of pH 12.00, and explain the difference.

When concentrated NaOH is added a little at a time, a white precipitate forms and then dissolves. Explain.

10.0 mL of 2.0 × 10⁻⁴ mol L⁻¹ ZnSO₄ is mixed with 10.0 mL of NaOH of pH 9.00. Will a precipitate form?

Show answer

1. Zn(OH)₂(s) ⇌ Zn²⁺(aq) + 2OH⁻(aq); Ks = [Zn²⁺][OH⁻]² (Achieved). 2. Ks = 4s³, so s = ∛(3.0 × 10⁻¹⁷ ÷ 4) = ∛(7.5 × 10⁻¹⁸) = 1.96 × 10⁻⁶ mol L⁻¹ (Merit). 3. [OH⁻] = 1.0 × 10⁻² mol L⁻¹, so s = 3.0 × 10⁻¹⁷ ÷ (1.0 × 10⁻²)² = 3.0 × 10⁻¹³ mol L⁻¹. The added OH⁻ is a common ion: it shifts the equilibrium to the left, so far less dissolves (Merit for the value with the reason). 4. At first, OH⁻ raises Q above Ks, so Zn(OH)₂ precipitates. With excess OH⁻, the complex ion forms, Zn²⁺ + 4OH⁻ → [Zn(OH)₄]²⁻, which removes Zn²⁺; Q falls below Ks, the equilibrium shifts to the right, and the precipitate dissolves (Excellence). 5. After mixing, [Zn²⁺] = 1.0 × 10⁻⁴ and [OH⁻] = 1.0 × 10⁻⁵ ÷ 2 = 5.0 × 10⁻⁶ mol L⁻¹, so Q = 1.0 × 10⁻⁴ × (5.0 × 10⁻⁶)² = 2.5 × 10⁻¹⁵. Q is greater than Ks = 3.0 × 10⁻¹⁷, so a precipitate forms (Excellence for the full calculation with dilution).

Summary

Recap

A saturated solution is a dynamic equilibrium; Ks is the product of the ion concentrations, each raised to its coefficient.
AB: Ks = s². A₂B and AB₂: Ks = 4s³. Compare solubilities with s, not Ks, when the types differ.
A common ion lowers solubility; complex ions and acids reacting with basic anions raise it. Ks changes only with temperature.
Precipitation happens when Q, worked out after dilution, is greater than Ks.

Situation	Effect on solubility
Common ion added	decreases
Ion removed into a complex ion	increases
Acid added to a hydroxide or carbonate	increases
Q > Ks after mixing	precipitate forms

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