Conversion of Ksp to Solubility

Learning Objectives

  • Perform calculations converting Ksp to solubility.


How do you purify water?

Purification of water for drinking and other uses is a complicated process. Heavy metals need to be removed, a process accomplished by addition of carbonates and sulfates. Lead contamination can present major health problems, especially for younger children. Lead sulfates and carbonates are very insoluble, so will precipitate out of solution very easily.

Heavy metals can be removed by precipitation with carbonates and sulfates

Conversion of Ksp to Solubility

Solubility Product Constants (25°C)
Compound Ksp Compound Ksp
AgBr 5.0 × 10 -13 CuS 8.0 × 10 -37
AgCl 1.8 × 10 -10 Fe(OH) 2 7.9 × 10 -16
Al(OH) 3 3.0 × 10 -34 Mg(OH) 2 7.1 × 10 -12
BaCO 3 5.0 × 10 -9 PbCl 2 1.7 × 10 -5
BaSO 4 1.1 × 10 -10 PbCO 3 7.4 × 10 -14
CaCO 3 4.5 × 10 -9 PbI 2 7.1 × 10 -9
Ca(OH) 2 6.5 × 10 -6 PbSO 4 6.3 × 10 -7
Ca 3 (PO 4 ) 2 1.2 × 10 -26 Zn(OH) 2 3.0 × 10 -16
CaSO 4 2.4 × 10 -5 ZnS 3.0 × 10 -23

The known Ksp values from the Table above can be used to calculate the solubility of a given compound by following the steps listed below.

  1. Set up an ICE problem (Initial, Change, Equilibrium) in order to use the Ksp value to calculate the concentration of each of the ions.
  2. The concentration of the ions leads to the molar solubility of the compound.
  3. Use the molar mass to convert from molar solubility to solubility.


The Ksp of calcium carbonate is 4.5 × 10 -9 . We begin by setting up an ICE table showing the dissociation of CaCO 3 into calcium ions and carbonate ions. The variable  s will be used to represent the molar solubility of CaCO 3 . In this case, each formula unit of CaCO 3 yields one Ca 2+ ion and one CO 3 2− ion. Therefore, the equilibrium concentrations of each ion are equal to s .

& text{CaCO}_3(s) quad rightleftarrows quad text{Ca}^{2+}(aq)+ text{CO}_3^{2-}(aq) \text{Initial }(text{M}) & qquad qquad qquad qquad quad 0.00 qquad quad 0.00 \text{Change }(text{M}) & qquad qquad qquad qquad quad +s qquad quad +s \qquad text{Equilibrium }(text{M}) & qquad qquad qquad qquad qquad s qquad qquad s

The Ksp expression can be written in terms of  s and then used to solve for s .

K_{sp}&=[ text{Ca}^{2+}][ text{CO}_3^{2-}]=(s)(s)=s^2 \s&=sqrt{K_{sp}}=sqrt{4.5 times 10^{-9}}=6.7 times 10^{-5} text{M}

The concentration of each of the ions at equilibrium is 6.7 × 10 -5  M. We can use the molar mass to convert from molar solubility to solubility.

frac{6.7 times 10^{-5} cancel{text{mol}}}{text{L}} times frac{100.09 text{g}}{1 cancel{text{mol}}} = 6.7 times 10^{-3} text{g/L}

So the maximum amount of calcium carbonate that is capable of dissolving in 1 liter of water at 25°C is 6.7 × 10 -3  grams. Note that in the case above, the 1:1 ratio of the ions upon dissociation led to the Ksp being equal to s^2 . This is referred to as a formula of the type AB , where  A is the cation and  B is the anion. Now let’s consider a formula of the type AB_2 , such as Fe(OH) 2 . In this case the setup of the ICE table would look like the following:

& text{Fe(OH)}_2(s) quad rightleftarrows quad text{Fe}^{2+}(aq)+2 text{OH}^-(aq) \text{Initial }(text{M}) & qquad qquad qquad qquad qquad 0.00 qquad quad 0.00 \text{Change }(text{M}) & qquad qquad qquad qquad qquad +s qquad quad +2s \text{Equilibrium }(text{M}) & qquad qquad qquad qquad qquad quad s qquad qquad 2s

When the Ksp expression is written in terms of s , we get the following result for the molar solubility.

K_{sp}&=[ text{Fe}^{2+}][ text{OH}^-]^2=(s)(2s)^2=4s^3 \s&=sqrt [3]{frac{K_{sp}}{4}}=sqrt [3]{frac{7.9 times 10^{-16}}{4}}=5.8 times 10^{-6} text{M}

The Table below shows the relationship between Ksp and molar solubility based on the formula.

Compound Type Example Ksp Expression Cation Anion Ksp in Terms of s
AB CuS [Cu 2+ ][S 2− ] s s s^2
AB 2 or A 2 B Ag 2 CrO 4 [Ag + ] 2 [CrO 4 2− ] 2s s 4s^3
AB 3 or A 3 B Al(OH) 3 [Al 3+ ][OH ] 3 s 3s 27s^4
A 2 B 3 or A 3 B 2 Ba 3 (PO 4 ) 2 [Ba 2+ ] 3 [PO 4 3− ] 2 3s 2s 108s^5

The Ksp expressions in terms of  s can be used to solve problems in which the Ksp is used to calculate the molar solubility as in the examples above. Molar solubility can then be converted to solubility.

Summary

  • The process of determining solubilities using Ksp values is described.


Practice

Read the material at the link below and do the problems at the end.

http://www.tonywhiddon.org/lhs/apchemistry/studyguides/solubility/ksp.htm

Review

  1. What information is needed to carry out these calculations?
  2. What allows the calculation of molar solubility?
  3. How is solubility determined?






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