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# 19.4: Slightly Soluble Salts

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• define solubility product constants.
• write solubility product constant expressions.
• calculate solubility product constants.

## Vocabulary

• solubility product constant

## Introduction

Since equilibrium constant expressions omit solids, the equilibrium constant expressions for the dissolution of a salt will have a denominator of 1, so no denominator will be shown. Here are some examples with their equilibrium constant expressions.

$\mathrm{NaCl}_{(s)} \rightleftharpoons \mathrm{Na}^+_{(aq)} + \mathrm{Cl}^-_{(aq)}$
$K = [\mathrm{Na}^+][\mathrm{Cl}^-]$
$\mathrm{Zn(IO}_3)_{2(s)} \rightleftharpoons \mathrm{Zn}^{2+}_{(aq)} + 2 \ \mathrm{IO}{^{-}}_{3(aq)}$
$K = [\mathrm{Zn}^{2+}][\mathrm{IO}_3^-]^2$
$\mathrm{PbCl}_{2(s)} \rightleftharpoons \mathrm{Pb}^{2+}_{(aq)} + 2 \ \mathrm{Cl}^-_{(aq)}$
$K = [\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2$

In this final section of the chapter, we will look at solubility equilibria.

## Salts in an Equilibrium System

Consider the dissociation reaction for copper(II) hydroxide:

$\mathrm{Cu(OH)}_{2(s)} \rightleftharpoons \mathrm{Cu}^{2+}_{(aq)} + 2 \ \mathrm{OH}^-_{(aq)})$

The equilibrium constant for this reaction is very low with a value of $2.2 \times 10^{-20}$. A very small value for $K$ means that the reactant (the solid salt) is heavily favored, so very few $\mathrm{Cu}^{2+}$ and $\mathrm{OH}^-$ ions will actually be formed when $\mathrm{Cu(OH)}_2$ is allowed to dissolve in water. In other words, the solution will become saturated very quickly.

This type of equilibrium will be established for any saturated solution that also contains some of the solid salt. When writing the equilibrium expression for a dissociation reaction, the equilibrium constant is often given the subscript “sp,” which stands for solubility product. The solubility product constant $K_{sp}$ works the same as any other equilibrium constant, but because calculations for this type of reaction are often quite similar, they are frequently differentiated in this way.

Example:

Tooth enamel is composed of the compound hydroxyapatite, $\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH}$, which has a $K_{sp}$ value of $6.8 \times 10^{-37}$. Write the dissociation equation and comment on the value of the equilibrium constant.

Solution:

$\mathrm{Ca}_5(\mathrm{PO}_4)_3\mathrm{OH}_{(aq)} \rightleftharpoons \mathrm{Ca}_5(\mathrm{PO}_4)^{+}_{3(aq)} + \mathrm{OH}^-_{(aq)}$

The equilibrium constant is very small, so the reactants are heavily favored. This means that tooth enamel will not dissolve readily in water (definitely a good thing).

Example: Write the dissociation reaction and the solubility product constant expression for each of the following solids.

1. $\mathrm{PbSO}_{4(s)}$
2. $\mathrm{Al(OH)}_{3(s)}$

Solution:

1. $\mathrm{PbSO}_{4(s)} \rightleftharpoons \mathrm{Pb}^{2+}_{(aq)} + \mathrm{SO}^{2-}_{4(aq)} \ \ \ \ \ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{SO}_4^{2-}]$
2. $\mathrm{Al(OH)}_{3(s)} \rightleftharpoons \mathrm{Al}^{3+}_{(aq)} + 3 \ \mathrm{OH}^-_{(aq)} \ \ \ \ \ K_{sp} =[\mathrm{Al}^{3+}][\mathrm{OH}^-]^3$

## Calculating Ksp From Solubility

The $K_{sp}$ of a slightly soluble salt can be calculated from its solubility. Solubilities are usually given in grams/liter, or occasionally moles/liter. If the solubility is given in moles/liter, you can follow the same process as demonstrated below, except you would skip the first step.

Given the solubility of copper(I) bromide ($0.0287 \ \mathrm{grams/liter \ H}_2\text{O}$), the molarity of a saturated $\mathrm{CuBr}$ solution can be determined. From that value, we can find the concentrations of the individual ions.

$\text{CuBr dissolved} = \frac {0.0287 \ \text{g/L}} {143.5 \ \text{g/mol}} = 0.000200 \ \text{M} = 2.00 \times 10^{-4} \ \text{M}$
$\mathrm{CuBr}_{(s)} \rightleftharpoons \mathrm{Cu}^+_{(aq)} + \mathrm{Br}^-_{(aq)}$

In this case, one $\mathrm{Cu}^+$ ion and one $\mathrm{Br}^-$ ion is formed from each formula unit of $\mathrm{CuBr}$, so when $2.00 \times 10^{-4} \ \mathrm{mol/L}$ of $\mathrm{CuBr}$ dissolves, the concentration of the ions in solution will be $[\mathrm{Cu}^+] = 2.00 \times 10^{-4} \ \mathrm{mol/L}$ and $[\mathrm{Br}^-] = 2.00 \times 10^{-4} \ \mathrm{mol/L}$.

The $K_{sp}$ expression for $CuBr$ is $K_{sp} = [Cu^+][Br^-]$. Substituting the calculated values into the $K_{sp}$ expression will give:

$K_{sp} = [\mathrm{Cu}^+][\mathrm{Br}^-] = (2.00 \times 10^{-4})(2.00 \times 10^{-4}) = 4.00 \times 10^{-8}$.

The process is only slightly more complicated for salts that dissociate into more than two ions. Consider the salt calcium phosphate. The solubility of calcium phosphate is $5.10 \times 10^{-5} \ \;\mathrm{g/L}$. First, we convert the solubility in $\mathrm{grams/liter}$ to $\mathrm{moles/liter}$.

$\text{molarity} \ \mathrm{Ca}_3(\mathrm{PO}_4)_2 \ \text{dissolved} = \frac {5.10 \times 10^{-5} \ \mathrm{g/L}} {310. \ \text{g/mol}} = 1.67 \times 10^{-7} \ \text{M}$

Then, using the dissolving equation, we determine the molarity of the ions in solution.

$\mathrm{Ca}_3(\mathrm{PO}_4)_{2(s)} \rightleftharpoons 3 \ \mathrm{Ca}^{2+} + 2 \ \mathrm{PO}_4^{3-}$

For every $\mathrm{mol/L}$ of calcium phosphate that dissolves, there will be 3 times as many calcium ions and 2 times as many phosphate ions in solution. Therefore,

$[\mathrm{Ca}^{2+}] = (3)(1.67 \times 10^{-7} \ \text{M}) = 5.01 \times 10^{-7} \ \text{M}$
$[\mathrm{PO}_4^{3-}] = (2)(1.67 \times 10^{-7} \ \text{M}) = 3.34 \times 10^{-7} \ \text{M}$

We finish the calculation by writing the $K_{sp}$ expression and substituting the molarities of the ions into the expression and calculate.

$K_{sp} = [\mathrm{Ca}^{2+}]^3[\mathrm{PO}_4^{3-}]^2 = (5.01 \times 10^{-7})^3(3.34 \times 10^{-7})^2 = 1.4 \times 10^{-32}$

### Calculating Solubility From Ksp

Not only can $K_{sp}$ be calculated from the solubility, but the solubility can be calculated from the $K_{sp}$ value. Given the $K_{sp}$ for $\mathrm{AgBr}$ ($K_{sp} = 1.60 \times 10^-{10}$), we can use the dissociation equation and some algebra to calculate the concentrations of each ion in solution.

$\mathrm{AgBr}_{(s)} \rightleftharpoons \mathrm{Ag}^+ + \mathrm{Br}^-$

If we allow $x$ to represent the moles of $\mathrm{AgBr}$ that dissolves in one liter of water, $[\mathrm{Ag}^+] = x$ and $[\mathrm{Br}^-] = x$, since only one of each ion is produced per formula unit of $\mathrm{AgBr}$. We can then write the $K_{sp}$ expression for $\mathrm{AgBr}$, set it equal to the given $K_{sp}$ value, substitute the assigned variables into the equation, and solve.

$K_{sp} = [\mathrm{Ag}^+][\mathrm{Br}^-] = 1.60 \times 10^{-10}$
$(x)(x) = 1.60 \times 10^{-10}$
$x = 1.26 \times 10^{-5} \ \text{M}$

Because of how we defined $x$, we can see that the molarity of $\mathrm{AgBr}$ will be the same so the solubility of $\mathrm{AgBr}$ is $1.26 \times 10^{-5} \ \mathrm{M}$. To find the solubility of $\mathrm{AgBr}$ in $\mathrm{grams/liter}$, we simply need to multiply the molarity by the molar mass.

Solubility of $\mathrm{AgBr \ in \ g/L} = (1.26 \times 10^{-5} \ \mathrm{mol/L})(188 \ \mathrm{g/mol}) = 2.4 \times 10^{-3} \ \mathrm{g/L}$

This problem becomes slightly more difficult when the salt generates more than two ions. Let's try lead(II) fluoride. Here is the dissolving equation, the $K_{sp}$ expression, and the $K_{sp}$ value for $\mathrm{PbF}_2$:

$\mathrm{PbF}_{2(s)} \rightleftharpoons \mathrm{Pb}^{2+} + 2 \ \mathrm{F}^-$
$K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{F}^-]^2 = 4.0 \times 10^{-8}$

Once again, we let $x$ represent the solubility of $\mathrm{PbF}_2 \ \mathrm{in \ moles/liter}$. Therefore, the concentration of lead ions in solution will be $x$ and the concentration of fluoride ions in solution will be $2x$. Substitute these variables into the $K_{sp}$ expression and solve for $x$.

$[\mathrm{Pb}^{2+}][\mathrm{F}^-]^2 = 4.0 \times 10^{-8}$
$(x)(2x)^2 = 4.0 \times 10^{-8}$
$4x^3 = 4.0 \times 10^{-8}$
$x^3 = 1.0 \times 10^{-8}$
$x = 2.2 \times 10^{-3} \ \mathrm{mol/L}$

When we assigned the variables, $x$ represented the molarity of the lead ions in solution, but $x$ also represented the molarity of lead(II) fluoride that had dissolved, so this is the solubility in moles/liter. To get grams/liter, we multiply by the molar mass.

Solubility of $\mathrm{PbF}_2 = (2.2 \times 10^{-3} \ \mathrm{mol/L})(245 \ \mathrm{g/mol}) = 0.53 \ \mathrm{g/L}$

You should look back at the assignment of variables and determine the molarity of fluoride ions in solution. There will be questions where you are asked to find the concentrations of the ions in solution from $K_{sp}$. You would use this same process, but the desired answer may not always be equal to $x$.

## Lesson Summary

• Equilibrium constants for slightly soluble salts are called solubility product constants.

## Review Questions

1. What is the solubility product constant? Give an example.
2. Why is solubility considered a special case for chemical equilibria?
3. Nickel hydroxide is a slightly soluble salt. Its dissociation reaction is represented as: $\mathrm{Ni(OH)}_{2(s)} \rightleftharpoons \mathrm{Ni}^{2+}_{(aq)} + 2 \ \mathrm{OH}^-_{(aq)}$. Which of the following represents the solubility product constant expression, $K_{sp}$?
1. $K_{sp} = \frac {[\mathrm{Ni}^{2+}][\mathrm{OH}^-]} {[\mathrm{Ni(OH)}_2]}$
2. $K_{sp} = \frac {[\mathrm{Ni}^{2+}][\mathrm{OH}^-]^2} {[\mathrm{Ni(OH)}_2]}$
3. $K_{sp} = [\mathrm{Ni}^{2+}][\mathrm{OH}^-]$
4. $K_{sp} = [\mathrm{Ni}^{2+}][\mathrm{OH}^-]^2$
4. The $K_{sp}$ for $\mathrm{AgBr}$ is $5.0 \times 10^{-13}$. What is $[\mathrm{Ag}^+]$ at equilibrium?
1. $5.0 \times 10^{-13} \ \mathrm{mol/L}$
2. $7.1 \times 10^{-7} \ \mathrm{mol/L}$
3. $2.5 \times 10^{-13} \ \mathrm{mol/L}$
4. not enough information is given
5. The $K_{sp}$ for $\mathrm{PbF}_2$ is $3.60 \times 10^{-8}$. What is $[\mathrm{F}^-]$ at equilibrium?
1. $3.60 \times 10^{-8} \ \mathrm{mol/L}$
2. $3.33 \times 10^{-3} \ \mathrm{mol/L}$
3. $4.16 \times 10^{-3} \ \mathrm{mol/L}$
4. $2.08 \times 10^{-3} \ \mathrm{mol/L}$
6. Magnesium hydroxide is the acive component in milk of magnesia, a suspension used to cure indigestion. It has an equilibrium constant of $6.3 \times 10^{-10}$. Write the dissociation equation and comment on the value of the equilibrium constant.
7. Write the dissociation reactions for the following salts as well as the $K_{sp}$ expressions.
1. calcium fluoride
2. chromium(II) carbonate
3. arsenic(III) sulfide

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## Date Created:

Feb 23, 2012

Mar 26, 2015
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