The Solubility-Product Constant, $K$ _sp

Learning Goal 32
Set up an expression for the solubility product constant for a salt.

An equilibrium can exist between a partially soluble substance and its solution:

For example:

BaSO

₄ (s)

Ba²⁺ (aq) + SO₄^2- (aq)

When writing the equilibrium-constant expression for the dissolution of $BaSO$ ₄, we remember that the concentration of a solid is constant. The expression is therefore:

K = [Ba

²⁺][SO₄^2-]

$K = K$ _sp, the solubility-product constant.

K

_sp = [Ba²⁺][SO₄^2-]

This constant is the product of the concentration of the ions involved in the equilibrium, raised to the powers of their coefficients in the equilibrium equation.

Click here or on the button on the menu bar for a table of K_sp values for some common salts.

Solubility and $K$ _sp

Three important definitions:

solubility: quantity of a substance that dissolves to form a saturated solution
molar solubility: the number of moles of the solute that dissolves to form a liter of saturated solution
$K$ _sp (solubility product): the equilibrium constant for the equilibrium between an ionic solid and its saturated solution

Learning Goal 33
Calculate $K$ _sp from solubility data and solubility from $K$ _sp.

EXAMPLE:

The $K$ _sp for $CaF$ ₂ is $3.9$ x $10$ ^-11. What is its solubility in water, in grams per liter?

We begin by writing the equilibrium involved:

CaF

₂ (s)

Ca²⁺ (aq) + 2F^- (aq)

We will say that $x$ equals the molar solubility. This means the concentrations of $Ca$ ²⁺ and $F$ ^- will be $x$ and 2 $x$ respectively.

K

_sp = [Ca²⁺][F^-]² = 3.9 x

10

^-11

3.9

10

^-11 = x(2x)² = 4x³

x = 2.1

10

^-4 M

This is the molar solubility. However, we want the solubility in grams per liter.

2.1

10

^-4 M x

78.1 g/mole = 1.6

10

^-2 g/L

The Common Ion Effect

Learning Goal 34
Calculate the effect of an added common ion on the solubility of a salt which is slightly soluble.

In a solubility equilibrium such as this,

CaF

₂(s)

Ca²⁺ (aq) + 2F^- (aq)

adding either $Ca$ ²⁺ or $F$ ^- will shift the equilibrium to the left, reducing the solubility of $CaF$ ₂.

EXAMPLE:

Find the molar solubility of $CaF$ ₂ (K_sp = 3.9 x $10$ ^-11) in a solution containing 0.010 M $Ca(NO$ ₃)₂.

Again, let $x$ represent the molar solubility of $CaF$ ₂. The concentration of $F$ ^- will be 2 $x$ , but this time the concentration of $Ca$ ²⁺ will be $x$ plus the addition of 0.010 M that are present.

Thus:

K

_sp = 3.9 x

10

^-11 = [Ca²⁺][F^-]² = (x + 0.010)(2x)²

Here, we assume that $x$ will be very small, because $K$ _sp is so small. We therefore simplify the equation as follows:

3.9

10

^-11 = (0.010)(2x)²

x

² = { 3.9 x

10

^-11 / 4(0.010) } = 9.8 x

10

^-10

x = 3.1

10

^-5 M

The Solubility-Product Constant, Ksp

Solubility and Ksp

EXAMPLE:

The Common Ion Effect

EXAMPLE:

The Solubility-Product Constant, $K$ _sp

Solubility and $K$ _sp