# 15.11: The Solubility-Product Constant

- Page ID
- 47588

Learning Objectives

- Understand the basic of solution-solid equilibria

We will now return to an important mathematical relationship we first learned about in our unit on Equilibrium, the equilibrium constant expression. Recall that for any general reaction:

\[aA + bB \rightleftharpoons cD + dD\]

an equilibrium constant expression can be defined as:

\[ K_{eq} = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}\]

Since saturated solutions are equilibrium systems, we can apply this mathematical relationship to solutions. We will refer to our equilibrium constant as \(K_{sp}\), where "sp" stands for "**s**olubility **p**roduct"

For our silver sulfate saturated solution,

\[ Ag_2SO_{4 (s)} \rightleftharpoons 2Ag^+_{(aq)} + SO^{2-}_{4(aq)} \]

we can write a solubility product constant expression as

\[ K_{sp} = \dfrac{ [Ag^+_{(aq)}]^2[SO_{4(aq)}^{2-}] }{ [Ag_2SO_{4(s)}] } \]

However, remember from our earlier introduction to the equilibrium constant expression that the concentrations of solids and liquids are * NOT* included in the expression because while their

*amounts*will change during a reaction, their

*concentrations*will remain constant. Therefore, we will write our solubility product constant expression for our saturated silver sulfate solution as:

\[K_{sp}= [Ag^+]^2[SO_4^{2-}]\]

Please note that it is **VERY IMPORTANT** to include the ion charges when writing this equation.

Example \(\PageIndex{1}\)

Write the expression for the solubility product constant, K_{sp}, for Ca_{3}(PO_{4})_{2}.

**Solution**

Step 1: Begin by writing the balanced equation for the reaction. Remember that polyatomic ions remain together as a unit and do not break apart into separate elements.

\[ Ca_3(PO_4)_{2 (s)} \rightleftharpoons 3 Ca^{2+}_{(aq)} + 2 PO^{3-}_{4(aq)} \]

Step 2: Write the expression for \(K_{sp}\):

\[K_{sp}= [Ca^{2+}]^3[PO_4^{3-}]^2\]

Solubility Product Tables that give K_{sp} values for various ionic compounds are available. Because temperature affects solubility, values are given for specific temperatures (usually 25°C). Recall what we learned about K_{eq}

- If K
_{eq}is very large, the concentration of the products must be much greater than the concentration of the reactants. The reaction essentially "goes to completion", which all - or most of - the reactants being used up to form the products. - If K
_{eq}is very small, the concentration of the reactants is much greater than the concentration of the products. The reaction does not occur to any great extent - most of the reactants remain unchanged, and there are few products produced. - When K
_{eq}is not very large or very small, then roughly equal amounts of reactants and products are present at equilibrium.

\(K_{sp}\), which again is just a special case of \(K_{eq}\), provides us with the same useful information:

A low value of K_{sp} indicates a substance that has a low solubility (it is generally insoluble); for ionic compounds this means that there will be few ions in solution.

- Iron(II) sulfide, FeS, is an example of a low K
_{sp}: K_{sp}= 4 ×10^{-19.}In a saturated solution of FeS there would be few Fe^{2+}or S^{2-}ions.

A large value of K_{sp} indicates a soluble substance; for ionic compounds it tells us that there will be many ions in solution.

- An example of a relatively large K
_{sp}would be for lead(II) chloride, PbCl_{2}which has a K_{sp}of 1.8 ×10^{-4}. A saturated solution of PbCl_{2}would have a relatively high concentration of Pb^{2+}and Cl^{-}ions.

There are several types of problems we can solve:

Example \(\PageIndex{1}\): Calculating K_{sp} of a Saturated Solution

The concentration of a saturated solution of BaSO_{4} is 3.90 × 10^{-5}M. Calculate K_{sp} for barium sulfate at 25°C

**Solution**

Always begin problems involving K_{sp} by writing a balanced equation:

BaSO_{4 (s)} Ba^{2+}_{(aq)} + SO_{4}^{2-}_{ (aq)}

Next, write the K_{sp} expression:

K_{sp}= [Ba^{2+}][SO_{4}^{2-}]

The question provides us with the concentration of the solution, BaSO_{4} . We need to find the concentration of the individual * ions* for our equation.

Recall from Section 2.5 - since 1 mole of BaSO_{4} produces 1 mole of Ba^{2+} and also 1 mole of SO_{4}^{2-}, then . . .

- [BaSO
_{4}] = 3.90 × 10^{-5}M - [Ba
^{2+}] = 3.90 × 10^{-5}M - [SO
_{4}^{2-}] = 3.90 × 10^{-5}M

Substitute values into the K_{sp} expression and solve for the unknown:

\[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \]

\[ = (3.90 \times 10^{-5})(3.90 \times 10^{-5}) \]

\[ = 1.52 \times 10^{-9} \]

Example \(\PageIndex{3}\): Calculating ion concentrations when K_{sp} is known

The \(K_{sp}\) for MgCO_{3} at 25°C is 2.0 × 10^{-8}. What are the ion concentrations in a saturated solution at this temperature?

**Solution**

As always, begin with a balanced equation:

MgCO_{3(s)} Mg^{2+}_{(aq)} + CO_{3}^{2-}_{(aq)}

Write the K_{sp} expression:

\[K_{sp}= [Mg^{2+}][CO_3^{2-}]\]

For this example, we are given the value for K_{sp} and need to find the ion concentrations.

We will let our unknown ion concentrations equal * x*.

**The balanced equation tells us that both Mg ^{2+} and CO_{3}^{2-} will have the same concentration!**

Substitute values into the equation and solve for the unknown

K_{sp} = [Mg^{2+}][CO_{3}^{2-}]

* x^{2}* = 2.0 × 10

^{-8}

* x* = √(2.0 × 10

^{-8})

find the square root of 2.0 × 10^{-8}

= 1.4× 10^{-4}M

*x = *[Mg^{2+}]

= 1.4× 10^{-4}M

**Answer**

AND * x = *[CO_{3}^{2-}]

= 1.4× 10^{-4}M