How to Calculate the Molar Solubility
- 1). Write a balanced equilibrium expression for the dissolution of the compound whose molar solubility will be calculated. For example, the equilibrium expression for calcium hydroxide, Ca(OH)2, would resemble Ca(OH)2 <---> Ca(2+) + 2 OH(-).
- 2). Find the Ksp value for the compound in question. Numerous websites, such as those provided in the Resources section, provide this information in tabulated form. In this case, calcium hydroxide exhibits a Ksp of 7.9 x 10^-6.
- 3). Write a Ksp expression for the equilibrium reaction written in step 1. For a generic compound AnBm, where n and m represents the subscripts of elements A and B, the equilibrium expression takes the form AnBm <---> nA(m+) + mB(n-). The Ksp expression then becomes Ksp = [A(m+)]^n * [B(n-)]^m (see references 3, slide 6), where square brackets denote concentration in units of moles per liter and the hat symbol, ^, represents an exponent. Thus, for the expression in step 1 for Ca(OH)2, Ksp = [Ca(2+)] * [OH(-)]^2.
- 4). Assign an equilibrium value of "x" to one of the ions. Technically, you can assign this value to any of the ions, but chemists usually assign this value to the ion with the lowest coefficient in the balanced equation. In the case of calcium hydroxide, Ca(2+) exhibits a coefficient of one and therefore x represents the concentration of Ca(2+). According to the balanced equation in step 1, the dissolution of calcium hydroxide produces two hydroxide ions for every calcium ion. As such, the concentration of OH(-) becomes "2x."
- 5). Substitute the "x" values from step 4 into the Ksp expression from step 3 and set this expression equal to the Ksp value found in step 2. For calcium hydroxide, Ksp = [Ca(2+)] * [OH(-)]^2 = x * (2x)^2 = 7.9 x 10^-6. Simplifying this expression gives 4x^3 = 7.9 x 10^-6.
- 6). Solve the expression from step 5 for x. In the example from step 5, first divide both sides by 4 to give x^3 = 2.0 x 10^-6. Then take the cube root of both sides to give x = 0.013.
- 7). Determine the molar solubility by establishing the ratio between the coefficients of the ion that was represented with x and the starting compound in the equilibrium expression from step 1, then multiply x by this ratio. For the example presented here, x represented Ca(2+), and the the coefficients of Ca(2+) and Ca(OH)2 were both 1. Therefore, the molar ratio between Ca(OH)2 and Ca(2+) is 1:1 or 1/1. The molar solubility of Ca(OH)2 is consequently 0.013 * 1/1 = 0.013 moles per liter.
Source...