# 19.13: Conversion of $$K_\text{sp}$$ to Solubility

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.

Compound $$K_\text{sp}$$ Compound $$K_\text{sp}$$
Table 19.13.1: Solubility Product Constants $$\left( 25^\text{o} \text{C} \right)$$
$$\ce{AgBr}$$ $$5.0 \times 10^{-13}$$ $$\ce{CuS}$$ $$8.0 \times 10^{-37}$$
$$\ce{AgCl}$$ $$1.8 \times 10^{-10}$$ $$\ce{Fe(OH)_2}$$ $$7.9 \times 10^{-16}$$
$$\ce{Al(OH)_3}$$ $$3.0 \times 10^{-34}$$ $$\ce{Mg(OH)_2}$$ $$7.1 \times 10^{-12}$$
$$\ce{BaCO_3}$$ $$5.0 \times 10^{-9}$$ $$\ce{PbCl_2}$$ $$1.7 \times 10^{-5}$$
$$\ce{BaSO_4}$$ $$1.1 \times 10^{-10}$$ $$\ce{PbCO_3}$$ $$7.4 \times 10^{-14}$$
$$\ce{CaCO_3}$$ $$4.5 \times 10^{-9}$$ $$\ce{PbI_2}$$ $$7.1 \times 10^{-9}$$
$$\ce{Ca(OH)_2}$$ $$6.5 \times 10^{-6}$$ $$\ce{PbSO_4}$$ $$6.3 \times 10^{-7}$$
$$\ce{Ca_3(PO_4)_2}$$ $$1.2 \times 10^{-26}$$ $$\ce{Zn(OH)_2}$$ $$3.0 \times 10^{-16}$$
$$\ce{CaSO_4}$$ $$2.4 \times 10^{-5}$$ $$\ce{ZnS}$$ $$3.0 \times 10^{-23}$$

The known $$K_\text{sp}$$ 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 $$K_\text{sp}$$ 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 $$K_\text{sp}$$ of calcium carbonate is $$4.5 \times 10^{-9}$$. We begin by setting up an ICE table showing the dissociation of $$\ce{CaCO_3}$$ into calcium ions and carbonate ions. The variable $$s$$ will be used to represent the molar solubility of $$\ce{CaCO_3}$$. In this case, each formula unit of $$\ce{CaCO_3}$$ yields one $$\ce{Ca^{2+}}$$ ion and one $$\ce{CO_3^{2-}}$$ ion. Therefore, the equilibrium concentrations of each ion are equal to $$s$$.

$\begin{array}{r|ccccc} & \ce{CaCO_3} \left( s \right) & \rightleftharpoons & \ce{Ca^{2+}} \left( aq \right) & + & \ce{CO_3^{2-}} \left( aq \right) \\ \hline \text{Initial} \: \left( \text{M} \right) & & & 0.00 & & 0.00 \\ \text{Change} \: \left( \text{M} \right) & & & +s & & +s \\ \text{Equilibrium} \: \left( \text{M} \right) & & & s & & s \end{array}$

The $$K_\text{sp}$$ expression can be written in terms of $$s$$ and then used to solve for $$s$$.

\begin{align} K_\text{sp} &= \left[ \ce{Ca^{2+}} \right] \left[ \ce{CO_3^{2-}} \right] = \left( s \right) \left( s \right) = s^2 \\ s &= \sqrt{K_\text{sp}} = \sqrt{4.5 \times 10^{-9}} = 6.7 \times 10^{-5} \: \text{M} \end{align}

The concentration of each of the ions at equilibrium is $$6.7 \times 10^{-5} \: \text{M}$$. 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 $$25^\text{o} \text{C}$$ is $$6.7 \times 10^{-3}$$ grams. Note that in the case above, the 1:1 ratio of the ions upon dissociation led to the $$K_\text{sp}$$ being equal to $$s^2$$. This is referred to as a formula of the type $$\ce{AB}$$, where $$\ce{A}$$ is the cation and $$\ce{B}$$ is the anion. Now let's consider a formula of the type $$\ce{AB_2}$$, such as $$\ce{Fe(OH)_2}$$. In this case the setup of the ICE table would look like the following:

$\begin{array}{r|ccccc} & \ce{Fe(OH)_2} \left( s \right) & \rightleftharpoons & \ce{Fe^{2+}} \left( aq \right) & + & 2 \ce{OH^-} \left( aq \right) \\ \hline \text{Initial} \: \left( \text{M} \right) & & & 0.00 & & 0.00 \\ \text{Change} \: \left( \text{M} \right) & & & +s & & +2s \\ \text{Equilibrium} \: \left( \text{M} \right) & & & s & & 2s \end{array}$

When the $$K_\text{sp}$$ expression is written in terms of $$s$$, we get the following result for the molar solubility.

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

The table below shows the relationship between $$K_\text{sp}$$ and molar solubility based on the formula.

Compound Type Example $$K_\text{sp}$$ Expression Cation Anion $$K_\text{sp}$$ in Terms of $$s$$
Table 19.13.2
$$\ce{AB}$$ $$\ce{CuS}$$ $$\left[ \ce{Cu^{2+}} \right] \left[ \ce{S^{2-}} \right]$$ $$s$$ $$s$$ $$s^2$$
$$\ce{AB_2}$$ or $$\ce{A_2B}$$ $$\ce{Ag_2CrO_4}$$ $$\left[ \ce{Ag^+} \right]^2 \left[ \ce{CrO_4^{2-}} \right]$$ $$2s$$ $$s$$ $$4s^3$$
$$\ce{AB_3}$$ or $$\ce{A_3B}$$ $$\ce{Al(OH)_3}$$ $$\left[ \ce{Al^{3+}} \right] \left[ \ce{OH^-} \right]^3$$ $$s$$ $$3s$$ $$27s^4$$
$$\ce{A_2B_3}$$ or $$\ce{A_3B_2}$$ $$\ce{Ba_3(PO_4)_2}$$ $$\left[ \ce{Ba^{2+}} \right]^3 \left[ \ce{PO_4^{3-}} \right]^2$$ $$3s$$ $$2s$$ $$108s^5$$

The $$K_\text{sp}$$ expressions in terms of $$s$$ can be used to solve problems in which the $$K_\text{sp}$$ 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 $$K_\text{sp}$$ values is described.

### Contributors

• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.