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Conversion of Ksp to Solubility

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Conversion of Ksp to Solubility
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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.

Conversion of  K_{sp} to Solubility

Solubility Product Constants (25°C)
Compound K_{sp} Compound K_{sp}
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  K_{sp} 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  K_{sp} 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  K_{sp} 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  K_{sp} 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  K_{sp} 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  K_{sp} 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  K_{sp} and molar solubility based on the formula.

Compound Type Example K_{sp}  Expression Cation Anion K_{sp} 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  K_{sp} expressions in terms of  s can be used to solve problems in which the  K_{sp} 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  K_{sp} 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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