Skip to content

Chemistry · Ch 6 — Equilibrium

Solubility Product Constant

6.13.1

Solubility Product Constant

The Solubility Product Constant

When a sparingly soluble ionic solid like barium sulphate is placed in water, it does not dissolve completely. Instead, an equilibrium is established between the undissolved solid and the ions present in the saturated solution. For barium sulphate, this equilibrium is written as:

BaSO4(s)⇌Ba2+(aq)+SO42−(aq)\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq)

The equilibrium constant for this heterogeneous equilibrium is:

K=[Ba2+][SO42−][BaSO4]K = \frac{[\text{Ba}^{2+}][\text{SO}_4^{2-}]}{[\text{BaSO}_4]}

Here's the key insight: the concentration of a pure solid is constant. It does not change as long as some solid is present. So we can absorb that constant into the equilibrium constant itself. Define a new constant:

Ksp=K[BaSO4]=[Ba2+][SO42−]K_{sp} = K[\text{BaSO}_4] = [\text{Ba}^{2+}][\text{SO}_4^{2-}]

This KspK_{sp} is called the solubility product constant, or simply the solubility product. It is the product of the concentrations of the ions in a saturated solution, each raised to the power of its stoichiometric coefficient in the dissolution equation.

Important

The solubility product constant is an equilibrium constant for the dissolution of a sparingly soluble salt. It applies only to saturated solutions in contact with undissolved solid.

At 298 K, the experimental value of KspK_{sp} for barium sulphate is 1.1×10−101.1 \times 10^{-10}. This means that in a saturated solution of BaSO4_4, the product [Ba2+][SO42−][\text{Ba}^{2+}][\text{SO}_4^{2-}] must equal 1.1×10−101.1 \times 10^{-10}.

Relating Solubility Product to Molar Solubility

For a salt like BaSO4_4 that dissociates into one cation and one anion, the concentrations of the two ions in the saturated solution are equal. If we let SS represent the molar solubility — the number of moles of salt that dissolve per litre of solution to form a saturated solution — then:

[Ba2+]=Sand[SO42−]=S[\text{Ba}^{2+}] = S \quad \text{and} \quad [\text{SO}_4^{2-}] = S

Substituting into the KspK_{sp} expression:

Ksp=(S)(S)=S2K_{sp} = (S)(S) = S^2

1.1×10−10=S21.1 \times 10^{-10} = S^2

S=1.1×10−10=1.05×10−5 mol L−1S = \sqrt{1.1 \times 10^{-10}} = 1.05 \times 10^{-5} \text{ mol L}^{-1}

So the molar solubility of barium sulphate in pure water at 298 K is 1.05×10−51.05 \times 10^{-5} mol L−1^{-1}.

General Formula for Salts with Different Ion Charges

Not all salts dissociate into ions with equal charges or in a 1:1 ratio. Consider a salt like zirconium phosphate, with the molecular formula Zr3(PO4)4\text{Zr}_3(\text{PO}_4)_4. It dissociates as:

Zr3(PO4)4(s)⇌3Zr4+(aq)+4PO43−(aq)\text{Zr}_3(\text{PO}_4)_4(s) \rightleftharpoons 3\text{Zr}^{4+}(aq) + 4\text{PO}_4^{3-}(aq)

If the molar solubility of this salt is SS, then from the stoichiometry:

[Zr4+]=3Sand[PO43−]=4S[\text{Zr}^{4+}] = 3S \quad \text{and} \quad [\text{PO}_4^{3-}] = 4S

The solubility product expression is:

Ksp=[Zr4+]3[PO43−]4=(3S)3(4S)4K_{sp} = [\text{Zr}^{4+}]^3[\text{PO}_4^{3-}]^4 = (3S)^3(4S)^4

Ksp=27S3×256S4=6912 S7K_{sp} = 27S^3 \times 256S^4 = 6912\,S^7

Therefore:

S=(Ksp6912)1/7S = \left(\frac{K_{sp}}{6912}\right)^{1/7}

Note

The exponent 7 comes from x+y=3+4=7x + y = 3 + 4 = 7, where xx and yy are the stoichiometric coefficients of the ions.

The General Case: Salt of Type Mx_xXy_y

For any sparingly soluble salt of the general formula Mx_xXy_y, the dissolution equilibrium is:

MxXy(s)⇌xMp+(aq)+yXq−(aq)\text{M}_x\text{X}_y(s) \rightleftharpoons x\text{M}^{p+}(aq) + y\text{X}^{q-}(aq)

where x×p+=y×q−x \times p^+ = y \times q^- to maintain electrical neutrality.

If the molar solubility is SS, then:

[Mp+]=xSand[Xq−]=yS[\text{M}^{p+}] = xS \quad \text{and} \quad [\text{X}^{q-}] = yS

The solubility product constant is:

Ksp=[Mp+]x[Xq−]y=(xS)x(yS)yK_{sp} = [\text{M}^{p+}]^x[\text{X}^{q-}]^y = (xS)^x(yS)^y

Ksp=xx⋅yy⋅S(x+y)K_{sp} = x^x \cdot y^y \cdot S^{(x+y)}

From this, we can solve for the molar solubility:

S(x+y)=Kspxx⋅yyS^{(x+y)} = \frac{K_{sp}}{x^x \cdot y^y}

S=(Kspxx⋅yy)1/(x+y)S = \left(\frac{K_{sp}}{x^x \cdot y^y}\right)^{1/(x+y)}

Ksp=xx⋅yy⋅S(x+y)K_{sp} = x^x \cdot y^y \cdot S^{(x+y)}

S=(Kspxx⋅yy)1/(x+y)S = \left(\frac{K_{sp}}{x^x \cdot y^y}\right)^{1/(x+y)}

The Reaction Quotient Qsp_{sp} and Precipitation …

Table 6.9The Solubility Product Constants, $K_{sp}$ of Some Common Ionic Salts at 298 K
SaltFormulaKspK_{sp}
Silver BromideAgBrAgBr5.0×10−135.0 \times 10^{-13}
Silver CarbonateAg2CO3Ag_2CO_38.1×10−128.1 \times 10^{-12}
Silver ChromateAg2CrO4Ag_2CrO_41.1×10−121.1 \times 10^{-12}
Silver ChlorideAgClAgCl1.8×10−101.8 \times 10^{-10}
Silver IodideAgIAgI8.3×10−178.3 \times 10^{-17}
Silver SulphateAg2SO4Ag_2SO_41.4×10−51.4 \times 10^{-5}
Aluminium HydroxideAl(OH)3Al(OH)_31.3×10−331.3 \times 10^{-33}
Barium ChromateBaCrO4BaCrO_41.2×10−101.2 \times 10^{-10}
Barium FluorideBaF2BaF_21.0×10−61.0 \times 10^{-6}
Barium SulphateBaSO4BaSO_41.1×10−101.1 \times 10^{-10}
Calcium CarbonateCaCO3CaCO_32.8×10−92.8 \times 10^{-9}
Calcium FluorideCaF2CaF_25.3×10−95.3 \times 10^{-9}
Calcium HydroxideCa(OH)2Ca(OH)_25.5×10−65.5 \times 10^{-6}
Calcium OxalateCaC2O4CaC_2O_44.0×10−94.0 \times 10^{-9}
Calcium SulphateCaSO4CaSO_49.1×10−69.1 \times 10^{-6}
Cadmium HydroxideCd(OH)2Cd(OH)_22.5×10−142.5 \times 10^{-14}
Cadmium SulphideCdSCdS8.0×10−278.0 \times 10^{-27}
Chromic HydroxideCr(OH)3Cr(OH)_36.3×10−316.3 \times 10^{-31}
Cuprous BromideCuBrCuBr5.3×10−95.3 \times 10^{-9}
Cupric CarbonateCuCO3CuCO_31.4×10−101.4 \times 10^{-10}
Cuprous ChlorideCuClCuCl1.7×10−61.7 \times 10^{-6}
Cupric HydroxideCu(OH)2Cu(OH)_22.2×10−202.2 \times 10^{-20}
Cuprous IodideCuICuI1.1×10−121.1 \times 10^{-12}
Cupric SulphideCuSCuS6.3×10−366.3 \times 10^{-36}
Ferrous CarbonateFeCO3FeCO_33.2×10−113.2 \times 10^{-11}
Ferrous HydroxideFe(OH)2Fe(OH)_28.0×10−168.0 \times 10^{-16}
Ferric HydroxideFe(OH)3Fe(OH)_31.0×10−381.0 \times 10^{-38}
Ferrous SulphideFeSFeS6.3×10−186.3 \times 10^{-18}
Mercurous BromideHg2Br2Hg_2Br_25.6×10−235.6 \times 10^{-23}
Mercurous ChlorideHg2Cl2Hg_2Cl_21.3×10−181.3 \times 10^{-18}
Mercurous IodideHg2I2Hg_2I_24.5×10−294.5 \times 10^{-29}
Mercurous SulphateHg2SO4Hg_2SO_47.4×10−77.4 \times 10^{-7}
Mercuric SulphideHgSHgS4.0×10−534.0 \times 10^{-53}
Magnesium CarbonateMgCO3MgCO_33.5×10−83.5 \times 10^{-8}
Magnesium FluorideMgF2MgF_26.5×10−96.5 \times 10^{-9}
Magnesium HydroxideMg(OH)2Mg(OH)_21.8×10−111.8 \times 10^{-11}
Magnesium OxalateMgC2O4MgC_2O_47.0×10−77.0 \times 10^{-7}
Manganese CarbonateMnCO3MnCO_31.8×10−111.8 \times 10^{-11}
Manganese SulphideMnSMnS2.5×10−132.5 \times 10^{-13}
Nickel HydroxideNi(OH)2Ni(OH)_22.0×10−152.0 \times 10^{-15}
Nickel SulphideNiSNiS4.7×10−54.7 \times 10^{-5}
Lead BromidePbBr2PbBr_24.0×10−54.0 \times 10^{-5}
Lead CarbonatePbCO3PbCO_37.4×10−147.4 \times 10^{-14}
Lead ChloridePbCl2PbCl_21.6×10−51.6 \times 10^{-5}
Lead FluoridePbF2PbF_27.7×10−87.7 \times 10^{-8}
Lead HydroxidePb(OH)2Pb(OH)_21.2×10−151.2 \times 10^{-15}
Lead IodidePbI2PbI_27.1×10−97.1 \times 10^{-9}
Lead SulphatePbSO4PbSO_41.6×10−81.6 \times 10^{-8}
Lead SulphidePbSPbS8.0×10−288.0 \times 10^{-28}
Stannous HydroxideSn(OH)2Sn(OH)_21.4×10−281.4 \times 10^{-28}
Start your 14-day free trial to unlock the full solution →
← Previous · 6.13Solubility Equilibria of Sparingly Soluble SaltsNext · 6.13.1 →Problem 6.26 — Calculate the solubility of A 2X3 in pure water, assuming that neither…