Equilibrium · Lesson 13 of 14

Solubility Products and Precipitation

“Use dissolution equilibria to calculate solubility and predict when a solid will form.”

Learning Objectives

• Relate molar solubility to the concentrations of ions using the salt formula. • Write Ksp and distinguish it from a current ion product. • Predict precipitation after accounting for dilution on mixing. • Explain common-ion and pH effects using coupled equilibria.

A sparingly soluble salt still establishes an equilibrium

Calcium chloride dissolves readily in water, whereas some other ionic solids dissolve only slightly. Dissolution separates lattice ions and surrounds them with solvent. Breaking the ionic lattice requires energy and hydration releases energy, but their enthalpy comparison alone cannot decide solubility: entropy also contributes to Gibbs energy. At equilibrium, a salt in contact with its saturated solution has a characteristic ion-activity product at the specified temperature.

The chapter uses three broad concentration categories: soluble above 0.1 M, slightly soluble between 0.01 and 0.1 M, and sparingly soluble below 0.01 M. These are convenient classifications rather than sharp physical boundaries. A sparingly soluble salt has a low dissolved concentration, not zero dissolved ions. Lattice strength, solvent identity, hydration and temperature all matter.

Barium sulfate dissolutionLaTeX

BaS O_{4} (s) ⇌ B a^{2 +} (aq) + S O_{4}^{2 -} (aq), K_{s p} \approx [B a^{2 +}] [S O_{4}^{2 -}]

The pure solid has unit activity and is omitted. The displayed ion concentrations are the dilute approximation to activities relative to their standard states.

Molar solubility S means the moles of salt formula units dissolved per litre of saturated solution. If BaSO4 dissolves without other important reactions and the ions are initially absent, both ion concentrations equal S. With the supplied Ksp of $1.1 \times 1 0^{- 10}$ , S is its square root, $1.05 \times 1 0^{- 5} M$ . A common ion or another reaction involving sulfate would alter this simple equality.

Stoichiometry determines the powers and concentration factors

A salt need not produce one ion of each kind. For A2X3, dissolving S moles per litre gives 2S of A3+ and 3S of X2−. Both the coefficients inside the concentrations and the powers in Ksp matter. Charges determine the neutral formula, but the exponents in the solubility expression are the stoichiometric coefficients, not simply the ionic charges.

General solubility relationLaTeX

M_{m} X_{n} (s) ⇌ m M^{p +} + n X^{q -}, K_{s p} = (m S)^{m} (n S)^{n} = m^{m} n^{n} S^{m + n}

Here mp = nq ensures charge balance. This simple S relation assumes no initial common ions, no significant hydrolysis, protonation or complex formation, and suitable dilute behaviour.

Recovering molar solubilityLaTeX

S = (\frac{K _{s p}}{m ^{m} n ^{n}})^{1/ (m + n)}

For zirconium phosphate, m = 3 and n = 4, so Ksp = 3^3 × 4^4 × S^7 = 6912 S^7.

Stoichiometry links solubility to ion concentrations— The salt is neutral because one 2+ cation accompanies two 1− anions. Ksp is [M2+][X−] squared, or 4S cubed in the simple model.

Example — A salt with a two-to-three ratio

Problem
Find the solubility of A2X3 when $K_{s p} = 1.1 \times 1 0^{- 23}$ and neither ion reacts significantly with water.

1.Dissociation gives $[A^{3 +}] = 2 S$ and $[X^{2 -}] = 3 S$ .
2.Then $K_{s p} = (2 S)^{2} (3 S)^{3} = 108 S^{5}$ .
3.Thus $S = (1.1 \times 1 0^{- 23} /108)^{1/5} = 1.004 \times 1 0^{- 5} M$ . Ion concentrations are approximately $2.01 \times 1 0^{- 5}$ and $3.01 \times 1 0^{- 5} M$ .

For an extremely insoluble hydroxide, water itself can supply more hydroxide than the formal value obtained from a bare stoichiometric S relation. Check that contribution before trusting the formula. For example, at maintained neutral pH at 298 K, hydroxide is about $1 0^{- 7}$ M; then Ksp directly constrains the free metal-ion concentration using that value. Hydrolysis or complex formation can further alter total dissolved metal. A formal S calculation is not automatically a complete physical model.

Example — Comparing salts with different formulas

Problem
Compare the molar solubilities of AgCN with $K_{s p} = 6.0 \times 1 0^{- 17}$ and Ni(OH)2 with $K_{s p} = 2.0 \times 1 0^{- 15}$ , using the simple concentration model.

1.For AgCN, $K_{s p} = S_{1}^{2}$ , so $S_{1} = 6.0 \times 1 0^{- 17} = 7.75 \times 1 0^{- 9} M$ .
2.For Ni(OH)2, $[N i^{2 +}] = S_{2}$ and $[O H^{-}] = 2 S_{2}$ , so $K_{s p} = 4 S_{2}^{3}$ .
3.Thus $S_{2} = (2.0 \times 1 0^{- 15} /4)^{1/3} = 7.94 \times 1 0^{- 6} M$ . Ni(OH)2 is more soluble in this model. Comparing Ksp values directly would not be a generally valid method for different salt stoichiometries.
4.The model neglects ion hydrolysis and other side equilibria. If they are important, total solubility must be obtained from all relevant equilibria rather than from Ksp alone.

Reference constants must be used with the correct salt formula

The following supplied 298 K data provide a reference for the calculations in this chapter. Each constant belongs to the specified solid and ionization stoichiometry. They are the chapter’s working values rather than a claim that every tabulated constant is universal across ionic strengths or solid forms. For example, mercury(I) salts contain the dimeric cation $H g_{2}^{2 +}$ , which must be treated as one ion in their dissociation equation.

Salt formula	Supplied Ksp at 298 K
AgBr	$5.0 \times 1 0^{- 13}$
Ag2CO3	$8.1 \times 1 0^{- 12}$
Ag2CrO4	$1.1 \times 1 0^{- 12}$
AgCl	$1.8 \times 1 0^{- 10}$
AgI	$8.3 \times 1 0^{- 17}$
Ag2SO4	$1.4 \times 1 0^{- 5}$
Al(OH)3	$1.3 \times 1 0^{- 33}$
BaCrO4	$1.2 \times 1 0^{- 10}$
BaF2	$1.0 \times 1 0^{- 6}$
BaSO4	$1.1 \times 1 0^{- 10}$
CaCO3	$2.8 \times 1 0^{- 9}$
CaF2	$5.3 \times 1 0^{- 9}$
Ca(OH)2	$5.5 \times 1 0^{- 6}$
CaC2O4	$4.0 \times 1 0^{- 9}$
CaSO4	$9.1 \times 1 0^{- 6}$
Cd(OH)2	$2.5 \times 1 0^{- 14}$
CdS	$8.0 \times 1 0^{- 27}$
Cr(OH)3	$6.3 \times 1 0^{- 31}$
CuBr	$5.3 \times 1 0^{- 9}$
CuCO3	$1.4 \times 1 0^{- 10}$
CuCl	$1.7 \times 1 0^{- 6}$
Cu(OH)2	$2.2 \times 1 0^{- 20}$
CuI	$1.1 \times 1 0^{- 12}$
CuS	$6.3 \times 1 0^{- 36}$
FeCO3	$3.2 \times 1 0^{- 11}$
Fe(OH)2	$8.0 \times 1 0^{- 16}$
Fe(OH)3	$1.0 \times 1 0^{- 38}$
FeS	$6.3 \times 1 0^{- 18}$
Hg2Br2	$5.6 \times 1 0^{- 23}$
Hg2Cl2	$1.3 \times 1 0^{- 18}$
Hg2I2	$4.5 \times 1 0^{- 29}$
Hg2SO4	$7.4 \times 1 0^{- 7}$
HgS	$4.0 \times 1 0^{- 53}$
MgCO3	$3.5 \times 1 0^{- 8}$
MgF2	$6.5 \times 1 0^{- 9}$
Mg(OH)2	$1.8 \times 1 0^{- 11}$
MgC2O4	$7.0 \times 1 0^{- 7}$
MnCO3	$1.8 \times 1 0^{- 11}$
MnS	$2.5 \times 1 0^{- 13}$
Ni(OH)2	$2.0 \times 1 0^{- 15}$
PbBr2	$4.0 \times 1 0^{- 5}$
PbCO3	$7.4 \times 1 0^{- 14}$
PbCl2	$1.6 \times 1 0^{- 5}$
PbF2	$7.7 \times 1 0^{- 8}$
Pb(OH)2	$1.2 \times 1 0^{- 15}$
PbI2	$7.1 \times 1 0^{- 9}$
PbSO4	$1.6 \times 1 0^{- 8}$
PbS	$8.0 \times 1 0^{- 28}$
Sn(OH)2	$1.4 \times 1 0^{- 28}$
SnS	$1.0 \times 1 0^{- 25}$
SrCO3	$1.1 \times 1 0^{- 10}$
SrF2	$2.5 \times 1 0^{- 9}$
SrSO4	$3.2 \times 1 0^{- 7}$
TlBr	$3.4 \times 1 0^{- 6}$
TlCl	$1.7 \times 1 0^{- 4}$
TlI	$6.5 \times 1 0^{- 8}$
ZnCO3	$1.4 \times 1 0^{- 11}$
Zn(OH)2	$1.0 \times 1 0^{- 15}$
ZnS	$1.6 \times 1 0^{- 24}$

Compare the current ion product with the solubility product

Use Qsp for the same ion-product expression evaluated with current free-ion activities or their concentration approximations. If Qsp exceeds Ksp, precipitation is thermodynamically favoured and can lower the ion product. If Qsp equals Ksp, a solution in contact with the solid is at saturation. If Qsp is below Ksp, more solid can dissolve if solid is available. An undersaturated solution with no solid present need not change by dissolution because there is nothing to dissolve.

Precipitation and dissolution— Use free-ion concentrations after mixing and dilution. Ksp stays fixed at fixed temperature while ion concentrations adjust.

Example — Mixing can produce a precipitate

Problem
Equal volumes of 0.0020 M AgNO3 and 0.0020 M NaCl are mixed. Predict precipitation using $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ .

1.The total volume doubles, so before precipitation each reacting ion has concentration 0.0010 M.
2.Calculate $Q_{s p} = (0.0010) (0.0010) = 1.0 \times 1 0^{- 6}$ .
3.Since this exceeds Ksp, AgCl precipitation is favoured. Do not use the unmixed 0.0020 M values; both concentrations change on dilution.

Common ions lower solubility in the simple model

Adding an ion that the dissolving salt already supplies increases Qsp and promotes precipitation or reduces additional dissolution. Passing HCl into saturated sodium chloride can precipitate high-purity NaCl because chloride activity increases. For such concentrated solutions, activities rather than bare molarities are essential. Similar equilibrium control is used to precipitate selected ions in gravimetric analysis, such as silver as AgCl, iron(III) as a hydroxide, or barium as BaSO4.

Example — Nickel hydroxide in sodium hydroxide

Problem
Find the molar solubility of Ni(OH)2 in 0.100 M NaOH using $K_{s p} = 2.0 \times 1 0^{- 15}$ and the concentration model.

1.If S dissolves, $[N i^{2 +}] = S$ while $[O H^{-}] = 0.100 + 2 S$ .
2.Write $2.0 \times 1 0^{- 15} = S (0.100 + 2 S)^{2}$ . Since the solubility is expected to be tiny, neglect 2S compared with 0.100.
3.Then $S = 2.0 \times 1 0^{- 15} /0.10 0^{2} = 2.0 \times 1 0^{- 13} M$ . The result makes 2S negligible, validating that approximation within the model.

Acid can dissolve salts by consuming their anions

For a salt MX whose anion X− is the conjugate base of a weak acid HX, added hydronium converts some free X− to HX. This lowers the free-ion product and permits more solid to dissolve. The total dissolved anion includes both X− and HX; Ksp uses only free X−. We must therefore satisfy the acid and solubility equilibria together.

Fraction of unprotonated dissolved anionLaTeX

f = \frac{[ X ^{-} ]}{[ X ^{-} ] + [ HX ]} = \frac{K _{a}}{K _{a} + [ H ^{+} ]}

This result follows from Ka = [H+][X−]/[HX]. It applies to the simple monoprotic anion model at a specified, maintained pH.

Solubility of a simple salt at fixed pHLaTeX

K_{s p} = S (f S), S = \frac{K _{s p}}{f} = K_{s p} \frac{K _{a} + [ H ^{+} ]}{K _{a}}

This formula is for a 1:1 salt MX with only the stated protonation side reaction, no initial common metal ion and ideal dilute behaviour. A lower pH can increase its total solubility.

Example — A pH effect on silver benzoate

Problem
For a simple silver-benzoate model, $K_{s p} = 2.5 \times 1 0^{- 13}$ and benzoic-acid $K_{a} = 6.46 \times 1 0^{- 5}$ . Compare solubility at maintained pH 3.19 with the unprotonated-anion estimate in pure water.

1.At pH 3.19, $[H^{+}] \approx 1 0^{- 3.19} = 6.46 \times 1 0^{- 4} M$ .
2.The free-benzoate fraction is approximately $f = 6.46 \times 1 0^{- 5} / (6.46 \times 1 0^{- 5} + 6.46 \times 1 0^{- 4}) = 1/11$ .
3.The unprotonated reference solubility is $S_{0} = K_{s p} = 5.0 \times 1 0^{- 7} M$ . At the specified pH, $S \approx S_{0} 11 = 1.66 \times 1 0^{- 6} M$ , about 3.32 times larger.

For an investigation, compare dissolution or precipitation of a sparingly soluble salt in water and in a solution containing one of its ions. Record the starting solutions, volumes and observed solid formation. Explain the observation by writing Qsp before the change and identifying how the common ion changes it. Concentration and pH changes must be recorded separately because a reagent can affect more than one equilibrium.

Common mistake

Ksp is not the same as molar solubility, and salts with different formulas cannot generally be ranked by Ksp alone. Use free-ion concentrations after dilution and include important side reactions when the simple model is insufficient.

Quick check

For CaF2 with molar solubility S and no side reactions, Ksp equals

Quick check

If Qsp is greater than Ksp, which change is favoured?

Quick check

Why does 0.10 M NaOH lower the simple-model solubility of Ni(OH)2?

Quick check

Before calculating Qsp after equal volumes are mixed, the original concentrations must generally be

Quick check

Lowering pH can increase solubility of a salt of a weak acid because

Quick check

Which quantity appears in Ksp for a salt whose anion can be protonated?

Practice Problems

Find the simple-model solubility and ion concentrations of Ag2CrO4 using Ksp = $1.1 \times 1 0^{- 12}$ .
Derive Ksp = 27S^4 for a salt M(OH)3 in pure water under the simple model.
Compare solubility of AgBr and Ag2CrO4 by calculating S, not merely comparing constants.
Equal volumes of 0.0020 M iodate and 0.0020 M copper(II) are mixed. Test for Cu(IO3)2 precipitation with Ksp = $7.4 \times 1 0^{- 8}$ .
Find the maximum free silver concentration compatible with 0.010 M chloride and Ksp(AgCl) = $1.8 \times 1 0^{- 10}$ .
Explain why acid can dissolve a carbonate or other salt with a basic anion, and state why a fixed-pH solubility formula requires the pH to be maintained.

Key Takeaways

A salt formula determines both ion-concentration factors and powers in Ksp. Compare Qsp with Ksp to predict saturation, dissolution or precipitation. Common ions suppress dissolution in the simple model, while protonating a basic anion can increase total solubility.

Previous · Lesson 12

Salt Hydrolysis and Buffer Solutions

Next · Lesson 14

Chapter Summary and Practice

A sparingly soluble salt still establishes an equilibrium

Stoichiometry determines the powers and concentration factors

Reference constants must be used with the correct salt formula

Compare the current ion product with the solubility product

Common ions lower solubility in the simple model

Acid can dissolve salts by consuming their anions

Quiz

Practice Problems

Key Takeaways