# 7.2: Ionic Equilibria between Solids and Solutions


Learning Objectives

• To calculate the solubility of an ionic compound from its Ksp

We begin our discussion of solubility and complexation equilibria—those associated with the formation of complex ions—by developing quantitative methods for describing dissolution and precipitation reactions of ionic compounds in aqueous solution. Just as with acid–base equilibria, we can describe the concentrations of ions in equilibrium with an ionic solid using an equilibrium constant expression.

## The Solubility Product

When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:

$Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)} \label{Eq1}$

As you will discover in more advanced chemistry courses, basic anions, such as S2−, PO43−, and CO32−, react with water to produce OH and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate.

The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product (Ksp) of the salt. Because the concentration of a pure solid such as Ca3(PO4)2 is a constant, it does not appear explicitly in the equilibrium constant expression. The equilibrium constant expression for the dissolution of calcium phosphate is therefore

$K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{Eq2a}$

$[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{Eq2b}$

At 25°C and pH 7.00, Ksp for calcium phosphate is 2.07 × 10−33, indicating that the concentrations of Ca2+ and PO43− ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of Ksp for some common salts are listed in Table $$\PageIndex{1}$$, which shows that the magnitude of Ksp varies dramatically for different compounds. Although Ksp is not a function of pH in Equations $$\ref{Eq2a}$$ and $$\ref{Eq2b}$$, changes in pH can affect the solubility of a compound as discussed later.

As with K, the concentration of a pure solid does not appear explicitly in Ksp.

Table $$\PageIndex{1}$$: Solubility Products for Selected Ionic Substances at 25°C
Solid Color Ksp Solid Color Ksp
*These contain the Hg22+ ion.
Acetates Iodides
Ca(O2CCH3)2·3H2O white 4 × 10−3 Hg2I2* yellow 5.2 × 10−29
Bromides PbI2 yellow 9.8 × 10−9
AgBr off-white 5.35 × 10−13 Oxalates
Hg2Br2* yellow 6.40 × 10−23 Ag2C2O4 white 5.40 × 10−12
Carbonates MgC2O4·2H2O white 4.83 × 10−6
CaCO3 white 3.36 × 10−9 PbC2O4 white 4.8 × 10−10
PbCO3 white 7.40 × 10−14 Phosphates
Chlorides Ag3PO4 white 8.89 × 10−17
AgCl white 1.77 × 10−10 Sr3(PO4)2 white 4.0 × 10−28
Hg2Cl2* white 1.43 × 10−18 FePO4·2H2O pink 9.91 × 10−16
PbCl2 white 1.70 × 10−5 Sulfates
Chromates Ag2SO4 white 1.20 × 10−5
CaCrO4 yellow 7.1 × 10−4 BaSO4 white 1.08 × 10−10
PbCrO4 yellow 2.8 × 10−13 PbSO4 white 2.53 × 10−8
Fluorides Sulfides
BaF2 white 1.84 × 10−7 Ag2S black 6.3 × 10−50
PbF2 white 3.3 × 10−8 CdS yellow 8.0 × 10−27
Hydroxides PbS black 8.0 × 10−28
Ca(OH)2 white 5.02 × 10−6 ZnS white 1.6 × 10−24
Cu(OH)2 pale blue 1 × 10−14
Mn(OH)2 light pink 1.9 × 10−13
Cr(OH)3 gray-green 6.3 × 10−31
Fe(OH)3 rust red 2.79 × 10−39

Solubility products are determined experimentally by directly measuring either the concentration of one of the component ions or the solubility of the compound in a given amount of water. However, whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, $$K_{sp}$$, like $$K$$, is defined in terms of the molar concentrations of the component ions.

A color photograph of a kidney stone, 8 mm in length. Kidney stones form from sparingly soluble calcium salts and are largely composed of Ca(O2CCO2)·H2O and Ca3(PO4)2. from Wikipedia.

Example $$\PageIndex{1}$$: Kidney Stones

Calcium oxalate monohydrate [Ca(O2CCO2)·H2O, also written as CaC2O4·H2O] is a sparingly soluble salt that is the other major component of kidney stones [along with Ca3(PO4)2]. Its solubility in water at 25°C is 7.36 × 10−4 g/100 mL. Calculate its Ksp.

Given: solubility in g/100 mL

Strategy:

1. Write the balanced dissolution equilibrium and the corresponding solubility product expression.
2. Convert the solubility of the salt to moles per liter. From the balanced dissolution equilibrium, determine the equilibrium concentrations of the dissolved solute ions. Substitute these values into the solubility product expression to calculate Ksp.

Solution

A We need to write the solubility product expression in terms of the concentrations of the component ions. For calcium oxalate monohydrate, the balanced dissolution equilibrium and the solubility product expression (abbreviating oxalate as ox2−) are as follows:

$$\mathrm{Ca(O_2CCO_2)}\cdot\mathrm{H_2O(s)}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{^-O_2CCO_2^-(aq)}+\mathrm{H_2O(l)}\hspace{5mm}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}][\mathrm{ox^{2-}}]$$

Neither solid calcium oxalate monohydrate nor water appears in the solubility product expression because their concentrations are essentially constant.

B Next we need to determine [Ca2+] and [ox2−] at equilibrium. We can use the mass of calcium oxalate monohydrate that dissolves in 100 mL of water to calculate the number of moles that dissolve in 100 mL of water. From this we can determine the number of moles that dissolve in 1.00 L of water. For dilute solutions, the density of the solution is nearly the same as that of water, so dissolving the salt in 1.00 L of water gives essentially 1.00 L of solution. Because each 1 mol of dissolved calcium oxalate monohydrate dissociates to produce 1 mol of calcium ions and 1 mol of oxalate ions, we can obtain the equilibrium concentrations that must be inserted into the solubility product expression. The number of moles of calcium oxalate monohydrate that dissolve in 100 mL of water is as follows:

$$\dfrac{7.36\times10^{-4}\textrm{ g}}{146.1\textrm{ g/mol}}=5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2)\cdot H_2O}$$

The number of moles of calcium oxalate monohydrate that dissolve in 1.00 L of the saturated solution is as follows:

$$\left(\dfrac{5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2\cdot)H_2O}}{\textrm{100 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1.00 L}}\right)=5.04\times10^{-5}\textrm{ mol/L}=5.04\times10^{-5}\textrm{ M}$$

Because of the stoichiometry of the reaction, the concentration of Ca2+ and ox2− ions are both 5.04 × 10−5 M. Inserting these values into the solubility product expression,

$K_{sp} = [Ca^{2+}][ox^{2−}] = (5.04 \times 10^{−5})(5.04 \times10^{−5}) = 2.54 \times 10^{−9} \nonumber$

In our calculation, we have ignored the reaction of the weakly basic anion with water, which tends to make the actual solubility of many salts greater than the calculated value.

Exercise $$\PageIndex{1}$$

One crystalline form of calcium carbonate (CaCO3) is the mineral sold as “calcite” in mineral and gem shops. The solubility of calcite in water is 0.67 mg/100 mL. Calculate its Ksp.

The reaction of weakly basic anions with $$H_2O$$ tends to make the actual solubility of many salts higher than predicted.

A crystal of calcite (CaCO3), illustrating the phenomenon of double refraction. When a transparent crystal of calcite is placed over a page, we see two images of the letters. Image used with permisison from Wikipedia

Calcite, a structural material for many organisms, is found in the teeth of sea urchins. The urchins create depressions in limestone that they can settle in by grinding the rock with their teeth. Limestone, however, also consists of calcite, so how can the urchins grind the rock without also grinding their teeth? Researchers have discovered that the teeth are shaped like needles and plates and contain magnesium. The concentration of magnesium increases toward the tip, which contributes to the hardness. Moreover, each tooth is composed of two blocks of the polycrystalline calcite matrix that are interleaved near the tip. This creates a corrugated surface that presumably increases grinding efficiency. Toolmakers are particularly interested in this approach to grinding.

Tabulated values of Ksp can also be used to estimate the solubility of a salt with a procedure that is essentially the reverse of the one used in Example $$\PageIndex{1}$$. In this case, we treat the problem as a typical equilibrium problem and set up a table of initial concentrations, changes in concentration, and final concentrations (ICE Tables), remembering that the concentration of the pure solid is essentially constant.

Example $$\PageIndex{2}$$

We saw that the Ksp for Ca3(PO4)2 is 2.07 × 10−33 at 25°C. Calculate the aqueous solubility of Ca3(PO4)2 in terms of the following:

1. the molarity of ions produced in solution
2. the mass of salt that dissolves in 100 mL of water at 25°C

Given: Ksp

Asked for: molar concentration and mass of salt that dissolves in 100 mL of water

Strategy:

1. Write the balanced equilibrium equation for the dissolution reaction and construct a table showing the concentrations of the species produced in solution. Insert the appropriate values into the solubility product expression and calculate the molar solubility at 25°C.
2. Calculate the mass of solute in 100 mL of solution from the molar solubility of the salt. Assume that the volume of the solution is the same as the volume of the solvent.

Solution:

1. A The dissolution equilibrium for Ca3(PO4)2 (Equation $$\ref{Eq2a}$$) is shown in the following ICE table. Because we are starting with distilled water, the initial concentration of both calcium and phosphate ions is zero. For every 1 mol of Ca3(PO4)2 that dissolves, 3 mol of Ca2+ and 2 mol of PO43− ions are produced in solution. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] will be +3x, and the change in [PO43−] will be +2x. We can insert these values into the table.

Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43−(aq)

$$\ce{Ca3(PO4)2}$$ $$[\ce{Ca^{2+}}]$$ $$\ce{[PO4^{3−}}]$$
initial pure solid 0 0
change +3x +2x
final pure solid 3x 2x

Although the amount of solid Ca3(PO4)2 changes as some of it dissolves, its molar concentration does not change. We now insert the expressions for the equilibrium concentrations of the ions into the solubility product expression (Equation 17.2):

\begin{align*}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2&=(3x)^3(2x)^2 \\2.07\times10^{-33}&=108x^5 \\1.92\times10^{-35}&=x^5 \\1.14\times10^{-7}\textrm{ M}&=x\end{align*}

This is the molar solubility of calcium phosphate at 25°C. However, the molarity of the ions is 2x and 3x, which means that [PO43−] = 2.28 × 10−7 and [Ca2+] = 3.42 × 10−7.

1. B To find the mass of solute in 100 mL of solution, we assume that the density of this dilute solution is the same as the density of water because of the low solubility of the salt, so that 100 mL of water gives 100 mL of solution. We can then determine the amount of salt that dissolves in 100 mL of water:

$$\left(\dfrac{1.14\times10^{-7}\textrm{ mol}}{\textrm{1 L}}\right)\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}} \right )\left(\dfrac{310.18 \textrm{ g }\mathrm{Ca_3(PO_4)_2}}{\textrm{1 mol}}\right)=3.54\times10^{-6}\textrm{ g }\mathrm{Ca_3(PO_4)_2}$$

Exercise $$\PageIndex{2}$$

The solubility product of silver carbonate (Ag2CO3) is 8.46 × 10−12 at 25°C. Calculate the following:

1. the molarity of a saturated solution
2. the mass of silver carbonate that will dissolve in 100 mL of water at this temperature

1.28 × 10−4 M

3.54 mg

## Relating Solubilities to Solubility Constants

The solubility (by which we usually mean the molar solubility) of a solid is expressed as the concentration of the "dissolved solid" in a saturated solution. In the case of a simple 1:1 solid such as AgCl, this would just be the concentration of Ag+ or Cl in the saturated solution. But for a more complicated stoichiometry such as as silver chromate, the solubility would be only one-half of the Ag+ concentration.

For example, let us denote the solubility of Ag2CrO4 as S mol L–1. Then for a saturated solution, we have

• $$[Ag^+] = 2S$$
• $$[CrO_4^{2–}] = S$$

Substituting this into Eq 5b above,

$(2S)^2 (S) = 4S^3 = 2.76 \times 10^{–12}$

$S= \left( dfrac{K_{sp}}{4} \right)^{1/3} = (6.9 \times 10^{-13})^{1/3} = 0.88 \times 10^{-4} \label{6a}$

thus the solubility is $$8.8 \times 10^{–5}\; M$$.

Note that the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction. For this reason it is meaningless to compare the solubilities of two salts having the formulas A2B and AB2, say, on the basis of their Ks values.

It is meaningless to compare the solubilities of two salts having different formulas on the basis of their Ks values.

Example $$\PageIndex{2}$$

The solubility of CaF2 (molar mass 78.1) at 18°C is reported to be 1.6 mg per 100 mL of water. Calculate the value of Ks under these conditions.

Solution

moles of solute in 100 mL; S = 0.0016 g / 78.1 g/mol = $$2.05 \times 10^{-5}$$ mol

\begin{align*} S &= \dfrac{2.05 \times 10^{ –5} mol}{0.100\; L} \\[4pt] &= 2.05 \times 10^{-4} M \end{align*}

\begin{align*}K_{sp} &= [Ca^{2+}][F^–]^2 \\[4pt] &= (S)(2S)^2 \\[4pt] &= 4 × (2.05 \times 10^{–4})^3 \\[4pt] &= 3.44 \times 10^{–11} \end{align*}

Example $$\PageIndex{3}$$

Estimate the solubility of La(IO3)3 and calculate the concentration of iodate in equilibrium with solid lanthanum iodate, for which Ks = 6.2 × 10–12.

Solution

The equation for the dissolution is

$\ce{La(IO_3)_3 <=> La^{3+ } + 3 IO3^{–}} \nonumber$

If the solubility is S, then the equilibrium concentrations of the ions will be

[La3+] = S and [IO3] = 3S. Then Ks = [La3+][IO3]3 = S(3S)3 = 27S4

27S4 = 6.2 × 10–12, S = ( ( 6.2 ÷ 27) × 10–12 )¼ = 6.92 × 10–4 M

[IO3] = 3S = 2.08 × 10–5 (M)

Example $$\PageIndex{4}$$: Cadmium

Cadmium is a highly toxic environmental pollutant that enters wastewaters associated with zinc smelting (Cd and Zn commonly occur together in ZnS ores) and in some electroplating processes. One way of controlling cadmium in effluent streams is to add sodium hydroxide, which precipitates insoluble Cd(OH)2 (Ks = 2.5E–14). If 1000 L of a certain wastewater contains Cd2+ at a concentration of 1.6E–5 M, what concentration of Cd2+ would remain after addition of 10 L of 4 M NaOH solution?

Solution

As with most real-world problems, this is best approached as a series of smaller problems, making simplifying approximations as appropriate.

Volume of treated water: 1000 L + 10 L = 1010 L

Concentration of OH on addition to 1000 L of pure water:

(4 M) × (10 L)/(1010 L) = 0.040 M

Initial concentration of Cd2+ in 1010 L of water:

$(1.6 \times 10^{–5}\; M) \left( \dfrac{100}{101} \right) \approx 1.6 \times 10^{–5}\; M$

The easiest way to tackle this is to start by assuming that a stoichiometric quantity of Cd(OH)2 is formed — that is, all of the Cd2+ gets precipitated.

Concentrations [Cd2+], M [OH], M
initial 1.6E–5 0.04
change –1.6E–5 –3.2E–5
final: 0 0.04 – 3.2E–5 ≈ .04

Now "turn on the equilibrium" — find the concentration of Cd2+ that can exist in a 0.04M OH solution:

Concentrations [Cd2+], M [OH], M
initial o 0.04
change +x +2x
at equilibrium x .04 + 2x.04

Substitute these values into the solubility product expression:

Cd(OH)2(s) = [Cd2+] [OH]2 = 2.5E–14

[Cd2+] = (2.5E–14) / (16E–4) = 1.6E–13 M

Note that the effluent will now be very alkaline:

$pH = 14 + log 0.04 = 12.6$

so in order to meet environmental standards an equivalent quantity of strong acid must be added to neutralize the water before it is released.

## Summary

The solubility product (Ksp) is used to calculate equilibrium concentrations of the ions in solution, whereas the ion product (Q) describes concentrations that are not necessarily at equilibrium. The equilibrium constant for a dissolution reaction, called the solubility product (Ksp), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, Ksp is defined in terms of the molar concentrations of the component ions.

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