# 15.11: The Solubility-Product Constant

- Page ID
- 47588

Learning Objectives

- Understand the basic concept of solution-solid equilibria.

We will now return to an important mathematical relationship that 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 "solubility product"

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 means that all—or most of—the reactants are 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

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

**Answer**

= 1.4× 10^{-4}M