How To Calculate Solubility From KSP

In chemistry, some ionic solids have low solubility in water. Some of the substance dissolves, and a lump of solid material remains. To calculate exactly how much dissolves, you use K_sp, the solubility product constant, along with an expression derived from the solubility equilibrium reaction for the substance.

Write the balanced solubility reaction equation for the substance you're interested in. This is the equation that describes what happens when the solid and dissolved parts reach equilibrium. To take an example, lead fluoride, PbF₂, dissolves into lead and fluoride ions in a reversible reaction:

\(\text{PbF}_2 ⇌ \text{Pb}^{2+} + 2\text{F}^-\)

Note that the positive and negative charges must balance on both sides. Also note, though lead has a +2 ionization, fluoride has −1. To balance the charges and account for the number of atoms for each element, you multiply the fluoride on the right side with the coefficient 2.

Look up the solubility product constant for the substance you're interested in. Chemistry books and websites have tables of ionic solids and their corresponding solubility product constants. To follow the example of lead fluoride, the K_sp is 3.7 × 10 ⁻⁸. This figure goes on the left side of the K_sp equation. On the right side, you break out each ion in square brackets. Note that a polyatomic ion would get its own brackets, you don't separate it out into individual elements. For the ions with coefficients, the coefficient becomes a power, as in the following expression:

\(\text{K}_\text{sp}= 3.7 × 10^{-8} = [\text{Pb}^{2+}][\text{F}^-]^2\)

The above expression equates the solubility product constant Ksp with the two dissolved ions but doesn't yet provide the concentration. To find the concentration, substitute X for each ion, as follows:

\(\text{K}_\text{sp}= 3.7 × 10^{-8} = (X)(X)^2\)

This treats each ion as distinct, both of which have a concentration molarity, and the product of those molarities equals K_sp, the solubility product constant. However, the second ion (F) is different. It has a coefficient of 2, which means each fluoride ion counts separately. To account for this after the substitution with X, put the coefficient inside the parenthesis:

\(\text{K}_\text{sp}= 3.7 × 10^{-8} = (X)(2X)^2\)

Now solve for X:

\(\begin{aligned}
3.7 × 10^{-8} &= (X)(4X^2)\)
\(3.7 × 10^{-8} &= 4X^3\)
\(X &= .0021 \text{ M}
\end{aligned}\)

This is the solution concentration in moles per liter.

To find the amount of the dissolved substance, multiply by liters of water, then multiply by the molar mass. For example, if your substance dissolved in 500 mL of water, 0.0021 moles per liter × 0.5 liters = 0.00105 moles. From the periodic table, the average atomic mass of lead is 207.2 and fluorine is 19.00. Since the lead fluoride molecule has 2 atoms of fluorine, multiply its mass by 2 to get 38.00. The total molar mass of lead fluoride is then 245.20 grams per mole. Since your solution has 0.0021 moles of dissolved substance, 0.0021 moles × 245.20 grams per mole = 0.515 grams of dissolved lead and fluoride ions.