# 14.10: The Solubility Product

In the section on precipitation reactions, we saw that there are some salts which dissolve in water to only a very limited extent. For example, if BaSO4 crystals are shaken with water, so little dissolves that it is impossible to see that anything has happened, as you will see in the video below. Nevertheless, the few Ba2+(aq) and SO42–(aq) ions that do go into solution increase the conductivity of the water, allowing us to measure their concentration. The video below shows the creation of Barium Sulfate in a precipitation reaction between barium chloride and sodium sulfate. Notice the white precipitate that forms, which is barium sulfate.

We find that at 25°C

$[\text{Ba}^{2+}]=\text{0.97} \times \text{10}^{-5}\text{ mol L}^{-1} = [\text{SO}_{4}^{2-}]\label{1}$

that we would describe the solubility of BaSO4 as 0.97 × 10–5 mol L–1 at this temperature. The solid salt and its ions are in dynamic equilibrium, and so we can write the equation

$\text{BaSO}_{4} ({s}) \rightleftharpoons \text{Ba}^{2+} ({aq}) + \text{SO}_{4}^{2-} ({aq})\label{2}$

As in other dynamic equilibria we have discussed, a particular Ba2+ ion will sometimes find itself part of a crystal and at other times find itself hydrated and in solution.

Since the concentration of BaSO4 has a constant value, it can be incorporated into Kc for Equation $$\ref{2}$$. This gives a special equilibrium constant called the solubility product Ksp:

$K_{sp}= K_{c}[\text{BaSO}_{4}] = [\text{Ba}^{2+}][\text{SO}_{4}^{2-}]\label{3}$

For BaSO4, Ksp is easily calculated from the solubility by substituting Equation $$\ref{1}$$ into $$\ref{3}$$:

\begin{align}K_{sp} & = (\text{0.97} \times \text{10}^{-5}\text{ mol L}^{-1}) (\text{0.97} \times \text{10}^{-5}\text{ mol L}^{-1})\\ &= \text{0.94} \times \text{10}^{-10}\text{ mol}^{2} \text{ L}^{-2}\end{align}

In the general case of an ionic compound whose formula is $$A_xB_y$$, the equilibrium can be written

$\text{A}_{x}\text{B}_{y} ({s}) \rightleftharpoons {x}\text{A}^{m+} ({aq}) + {y}\text{A}^{n+} ({aq})$

The solubility product is then

$\text{K}_{sp} = [\text{A}^{m+}]^{x}[\text{B}^{n+}]^{y}$

Solubility products for some of the more common sparingly soluble compounds are given in the table below.

Table $$\PageIndex{1}$$ ​Solubility Product Constants for Some Inorganic Compounds at 25 °C1
Substance Ks Substance Ksp
Aluminum Compounds Barium Compounds
AlAsO4 1.6 × 10-16 Ba3(AsO4)2 8.0 × 10-15
Al(OH)3 amorphous 1.3 × 10-33 BaCO3 5.1 × 10-9
AlPO4 6.3 × 10-19 BaC2O4 1.6 × 10-7
Bismuth Compounds BaCrO4 1.2 × 10-10
BiAsO4 4.4 ×10-10 BaF2 1.0 × 10-6
BiOCl2 7.0 × 10-9 Ba(OH)2 5 × 10-3
BiO(OH) 4 × 10-10 Ba3(PO4)2 3.4 × 10-23
Bi(OH)3 4 ×10-31 BaSeO4 3.5 × 10-8
Bil3 8.1 ×10-19 BaSO4 1.1 × 10-10
BiPO4 1.3 ×10-23 BaSO3 8 × 10-7
Cadmium Compounds BaS2O3 1.6 × 10-5
Cd3(AsO4)2 2.2 ×10-33 Calcium Compounds
CdCO3 5.2 ×10-12 Ca3(AsO4)2 6.8 ×10-19
Cd(CN)2 1.0 ×10-8 CaCO3 2.8 ×10-9
Cd2[Fe(CN)6] 3.2 ×10-17 CaCrO4 7.1 ×10-4
Cd(OH)2 fresh 2.5 ×10-14 CaC2O4 • H2O3 4 × 10-9
Chromium Compounds CaF2 5.3 ×10-9
CrAsO4 7.7 × 10-21 Ca(OH)2 5.5 ×10-6
Cr(OH)2 2 × 10-16 CaHPO4 1 × 10-7
Cr(OH)3 6.3 × 10-31 Ca3(PO4)2 2.0 × 10-29
CrPO4 • 4H2O green 2.4 × 10-23 CaSeO4 8.1 × 10-4
CrPO4 • 4H2O violet 1.0 × 10-17 CaSO4 9.1 × 10-6
Cobalt Compounds CaSO3 6.8 × 10-8
Co3(AsO4)2 7.6 × 10-29 Copper Compounds
CoCO3 1.4 × 10-13 CuBr 5.3 × 10-9
Co(OH)2 fresh 1.6 × 10-15 CuCl 1.2 × 10-6
Co(OH)3 1.6 × 10-44 CuCN 3.2 × 10-20
CoHPO4 2 × 10-7 CuI 1.1 × 10-12
CO3(PO4)2 2 × 10-35 CuOH 1 × 10-14
Gold Compounds CuSCN 4.8 × 10-15
AuCl 2.0 × 10-13 Cu3(AsO4)2 7.6 × 10-36
AuI 1.6 × 10-23 CuCO3 1.4 × 10-10
AuCl3 3.2 × 10-25 Cu2[Fe(CN)6] 1.3 × 10-16
Au(OH)3 5.5 × 10-46 Cu(OH)2 2.2 × 10-20
AuI3 1 × 10-46 Cu3(PO4)2 1.3 × 10-37
FeCO3 3.2 × 10-11 Pb3(AsO4)2 4.0 × 10-36
Fe(OH)2 8.0 × 10-16 PbBr2 4.0 × 10-5
FeC2O4 • 2H2O3 3.2 × 10-7 PbCO3 7.4 × 10-14
FeAsO4 5.7 × 10-21 PbCl2 1.6 × 10-5
Fe4[Fe(CN)6]3 3.3 × 10-41 PbCrO4 2.8 × 10-13
Fe(OH)3 4 × 10-38 PbF2 2.7 × 10-8
FePO4 1.3 × 10-22 Pb(OH)2 1.2 × 10-15
Magnesium Compounds PbI2 7.1 × 10-9
Mg3(AsO4)2 2.1 × 10-20 PbC2O4 4.8 × 10-10
MgCO3 3.5 × 10-8 PbHPO4 1.3 × 10-10
MgCO3 • 3H2O3 2.1 × 10-5 Pb3(PO4)2 8.0 × 10-43
MgC2O4 • 2H2O3 1 × 10-8 PbSeO4 1.4 × 10-7
MgF2 6.5 × 10-9 PbSO4 1.6 × 10-8
Mg(OH)2 1.8 × 10-11 Pb(SCN)2 2.0 × 10-5
Mg3(PO4)2 10-23 to 10-27 Manganese Compounds
MgSeO3 1.3 × 10-5 Mn3(AsO4)2 1.9 × 10-29
MgSO3 3.2 × 10-3 MnCO3 1.8 × 10-11
MgNH4PO4 2.5 × 10-13 Mn2[Fe(CN)6] 8.0 × 10-13
Mercury Compounds Mn(OH)2 1.9 × 10-13
Hg2Br2 5.6 × 10-23 MnC2O4 • 2H2O3 1.1 × 10-15
Hg2CO3 8.9 × 10-17 Nickel Compounds
Hg2(CN)2 5 × 10-40 Ni3(AsO4)2 3.1 × 10-26
Hg2Cl2 1.3 × 10-18 NiCO3 6.6 × 10-9
Hg2CrO4 2.0 × 10-9 2 Ni(CN)2 → Ni2+ + Ni(CN)42 1.7 × 10-9
Hg2(OH)2 2.0 × 10-24 Ni2[Fe(CN)6] 1.3 × 10-15
Hg2l2 4.5 × 10-29 Ni(OH)2 fresh 2.0 × 10-15
Hg2SO4 7.4 × 10-7 NiC2O4 4 × 10-10
Hg2SO3 1.0 × 10-27 Ni3(PO4)2 5 × 10-31
Hg(OH)2 3.0 × 10-26 Silver Compounds
Strontium Compounds Ag3AsO4 1.0 × 10-22
Sr3(AsO4)2 8.1 × 10-19 AgBr 5.0 × 10-13
SrCO3 1.1 × 10-10 Ag2CO3 8.1 × 10-12
SrCrO4 2.2 × 10-5 AgCl 1.8 × 10-10
SrC2O4 • H2O3 1.6 × 10-7 Ag2CrO4 1.1 × 10-12
Sr3(PO4)2 4.0 × 10-28 AgCN 1.2 × 10-16
SrSO3 4 × 10-8 Ag2Cr2O7 2.0 × 10-7
SrSO4 3.2 × 10-7 Ag4[Fe(CN)6] 1.6 × 10-41
Tin Compounds AgOH 2.0 × 10-8
Sn(OH)2 1.4 × 10-28 AgI 8.3 × 10-17
Sn(OH)4 1 × 10-56 Ag3PO4 1.4 × 10-16
Zinc Compounds Ag2SO4 1.4 × 10-5
Zn3(AsO4)2 1.3 × 10-28 Ag2SO3 1.5 × 10-14
ZnCO3 1.4 × 10-11 AgSCN 1.0 × 10-12
Zn2[Fe(CN)6] 4.0 × 10-16
Zn(OH)2 1.2 × 10-17
ZnC2O4 2.7 × 10-8
Zn3(PO4)2 9.0 × 10-33

1. Taken from Patnaik, Pradyot, Dean’s Analytical Chemistry Handbook, 2nd ed., New York: McGraw-Hill, 2004, Table 4.2 (published on the Web by Knovel, http://www.knovel.com).

2. Taken from Meites, L. ed., Handbook of Analytical Chemistry, 1st ed., New York: McGraw-Hill, 1963.

3. Because [H2O] does not appear in equilibrium constants for equilibria in aqueous solution in general, it does not appear in the Ksp expressions for hydrated solids.

No metal sulfides are listed in this table because sulfide ion is such a strong base that the usual solubility product equilibrium equation does not apply. See Myers, R. J. Journal of Chemical Education, Vol. 63, 1986; pp. 687-690.

Example $$\PageIndex{1}$$: Equilibrium

When crystals of PbCl2 are shaken with water at 25°C, it is found that 1.62 × 10–2 mol PbCl2 dissolves per cubic decimeter of solution. Find the value of Ksp at this temperature.

Solution

We first write out the equation for the equilibrium:

$\text{PbCl}_{2}({s})\rightleftharpoons \text{Pb}^{2+}({aq}) = \text{2Cl}^{-}({aq})$

so that

$\text{K}_{sp}\text{PbCl}_{2} = [\text{Pb}^{2+}][\text{Cl}^{-}]^{2}$

Since 1.62 × 10–2 mol PbCl2 dissolves per cubic decimeter, we have

$[\text{Pb}^{2+}]=\text{1.62} \times \text{10}^{-2} \text{mol L}^{-1}$

while

$[\text{Cl}^{-}]=\text{2} \times \text{1.62} \times \text{10}^{-2} \text{mol L}^{-1}$

since 2 mol Cl ions are produced for each mol PbCl2 which dissolves. Thus

\begin{align}{K}_{sp}= (\text{1.62}\times \text{10}^{-2}\text{mol L}^{-1})(\text{2 } \times \text{ 1.62 } \times \text{ 10}^{-2} \text{mol L}^{-1})\text{ }^{2}\\\text{ } = \text{1.70 } \times \text{ 10}^{-5} \text{ mol}^{3} \text{L} ^{-3}\end{align}

Example $$\PageIndex{2}$$: Solubility

The solubility product of silver chromate, Ag2CrO4, is 1.0 × 10–12 mol3 L–3. Find the solubility of this salt.

Solution

Again we start by writing the equation

$\text{Ag}_{2}\text{CrO } ({s}) \rightleftharpoons \text {2Ag}^{2+} ({aq}) + \text{CrO}_{4}^{2-} ({aq})$

from which

${K}_{sp}(\text{Ag}_{2}\text{CrO}_{4})= [\text{Ag}^{+}]^{2} [\text{CrO}_{4}^{2-}]= \text{1.0} \times \text{10}^{-12} \text{mol}^{3} \text{L}^{-3}$

Let the solubility be x mol L–1. Then

$[\text{CrO}_{4}^{2-}]= {x } \text{ mol } \text{L}^{-1}$

and

$[\text{Ag}^{+}] = 2x \text{mol L}^{-1}$

Thus

\begin{align}{K}_{sp}= (\text{2}{x} \text{ mol } \text{L}^{-1})^{2} {x} \text{ mol L}^{-1}\\\text{ }= (\text{2}{x})^{2} { x}\text{ mol}^{3} \text{ L}^{-3} = \text{1.0} \times \text{10}^{-12}\text{ mol}^{3} \text{ L}^{-3} \, \end{align}

or

$4x^{3} = \text{1.0 x 10}^{-12}$

and

$x^{\text{3}}=\frac{\text{1.0}}{\text{4}}\text{ }\times \text{ 10}^{12}=\text{2.5 }\times \text{ 10}^{-13}=\text{250 }\times \text{ 10}^{-15}$

so that

$x=\sqrt[\text{3}]{\text{250}}\text{ }\times \text{ }\sqrt[\text{3}]{\text{10}^{-\text{15}}}=\text{6}\text{.30 }\times \text{ 10}^{-\text{5}}$

Thus the solubility is 6.30 × 10–5 mol L–1.