The Solubility Product

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:56


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. 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 concentrationA measure of the ratio of the quantity of a substance to the quantity of solvent, solution, or ore. Also, the process of making something more concentrated.. We find that at 25°C


[\text{Ba}^{2+}]=\text{0.97} \times \text{10}^{-5}\text{ mol dm}^{-3} = [\text{SO}_{4}^{2-}]      (1)


so that we would describe the solubility of BaSO4 as 0.97 × 10–5 mol dm–3 at this temperatureA physical property that indicates whether one object can transfer thermal energy to another object.. The solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. salt and its ions are in dynamic equilibriumA self-establishing state in which the concentrations of reactant and product species remains constant forever; called dynamic because it is reached when opposing processes occur at the same rate; if a change in conditions causes a system not to be at equilibrium, the system will return to equilibrium in a way that partially counteracts the change in conditions., and so we can write the equation


\text{BaSO}_{4} ({s}) \rightleftharpoons \text{Ba}^{2+} ({aq}) +  \text{SO}_{4}^{2-} ({aq})       (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 Eq. (2). This gives a special equilibrium constantThe value of the equilibrium constant expression when equilibrium concentrations are substituted; a value greater than one indicates the position of equilibrium lies toward products (product-favored), and a value less than one indicates the position of equilibrium lies toward reactants (reactant-favored). called the solubility productThe equilibrium constant expression for the dissolution of an electrolyte; the reactant is a solid and its concentration does not appear in the expression, which is a product of the concentrations of the products (raised the to appropriate powers). Ksp:


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


For BaSO4, Ksp is easily calculated from the solubility by substituting Eq. (1) into (3):


\begin{align}K_{sp}&= (\text{0.97} \times \text{10}^{-5}\text{ mol dm}^{-3}) (\text{0.97} \times \text{10}^{-5}\text{ mol dm}^{-3})\
\text{ }&= \text{0.94} \times \text{10}^{-10}\text{ mol}^{2} \text{ dm}^{-6}\end{align}


APPENDIX H


Solubility Product Constants for Some InorganicPertaining to the chemistry of elements other than carbon and compounds containing at most a small amount of carbon. Compounds at 25 °C1

Substance Ksp 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
Iron Compounds Lead Compounds
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 metalAn element characterized by a glossy surface, high thermal and electrical conductivity, malleability, and ductility. sulfides are listed in this table because sulfide ion is such a strong baseA base that dissociates completely or ionizes completely in a particular solvent. that the usual solubility product equilibrium equation does not apply. See Myers, R. J. Journal of Chemical Education, Vol. 63, 1986; pp. 687-690.



In the general case of an ionic compoundA compound containing oppositely charged ions held together by electrostatic attraction. Usually the ions are in a crystal lattice with positive ions surrounded by negative ions and negative ions surrounded by positive ions. whose formula is AxBy, the equilibrium can be written


\text{A}_{x}\text{B}_{y} ({s}) \rightleftharpoons {x}\text{A}^{m+} ({aq}) + {y}\text{A}^{n+} ({aq})      (4)


The solubility product is then


{K}_{sp}= [\text{A}^{m+}]\text{ }^{x} [\text{B}^{n+}]\text{ }^{y}\,      (5)


Solubility products for some of the more common sparingly solubleAble to dissolve in a solvent to a significant extent. compounds are given in the table above.



EXAMPLE 1 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}) ...(6)


so that


             {K}_{sp}(\text{PbCl}_{2})= [\text{Pb}^{2+}] [\text{Cl}^{-}]^{2}\,...(7)


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 dm}^{-3}


while      [\text{Cl}^{-}]=\text{2} \times \text{1.62} \times \text{10}^{-2} \text{mol dm}^{-3}


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 dm}^{-3})(\text{2 } \times \text{ 1.62 } \times \text{ 10}^{-2} \text{mol dm}^{-3})\text{ }^{2}\
\text{ }& = \text{1.70 } \times \text{ 10}^{-5} \text{ mol}^{3} \text{dm} ^{-9}\end{align}



EXAMPLE 2 The solubility product of silver chromate, Ag2CrO4, is 1.0 × 10–12 mol3 dm–9. 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{dm}^{-9}


Let the solubility be x mol dm–3. Then


                 [\text{CrO}_{4}^{2-}]= {x }  \text{ mol } \text{dm}^{-3}

      and      [\text{Ag}^{+}] = \text{2}{x} \text{ mol } \text{dm}^{-3}\,


Thus


          \begin{align}{K}_{sp}&= (\text{2}{x} \text{ mol } \text{dm}^{-3})^{2} {x} \text{ mol dm}^{-3}\
\text{ }&= (\text{2}{x})^{2} { x}\text{ mol}^{3} \text{ dm}^{-9} = \text{1.0} \times \text{10}^{-12}\text{ mol}^{3} \text{ dm}^{-9} \, \end{align}


or              \text{4}{x}^{3} = \text{1.0} \times \text{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 dm–3.