How To Calculate The Molar Solubility

How to calculate the molar solubility is a fundamental skill for anyone studying chemistry, especially when dealing with sparingly soluble salts and equilibrium concepts. Worth adding: molar solubility tells us how many moles of a solid dissolve per liter of solution to reach saturation, and it is directly linked to the solubility product constant (Ksp) of the compound. Understanding this relationship allows students to predict whether a precipitate will form, to design quantitative analyses, and to interpret the effects of common ions, pH, and complexation on solubility. The following guide breaks down the theory, provides a clear step‑by‑step procedure, works through illustrative examples, and highlights common pitfalls to avoid.

What Is Molar Solubility?

Molar solubility, usually expressed in moles per liter (mol L⁻¹), is the concentration of a dissolved species at equilibrium with its solid phase. For a generic salt AB that dissociates as

[ \text{AB(s)} \rightleftharpoons \text{A}^+(aq) + \text{B}^-(aq) ]

the molar solubility (s) represents the amount of AB that dissolves to produce s mol L⁻¹ of A⁺ and s mol L⁻¹ of B⁻ (assuming a 1:1 stoichiometry). When the dissolution reaction involves different coefficients, the relationship between s and the ion concentrations must reflect those coefficients.

Relationship Between Molar Solubility and Ksp

The solubility product constant (Ksp) is an equilibrium expression that depends only on temperature. For the dissociation above,

[ K_{sp} = [\text{A}^+][\text{B}^-] ]

If the molar solubility is s, then at equilibrium

[ [\text{A}^+] = s \quad \text{and} \quad [\text{B}^-] = s ]

so

[ K_{sp} = s \times s = s^2 ]

For salts with different stoichiometries, the expression becomes more complex. To give you an idea, for A₂B dissociating as

[ \text{A}_2\text{B(s)} \rightleftharpoons 2\text{A}^+(aq) + \text{B}^{2-}(aq) ]

the equilibrium concentrations are

[ [\text{A}^+] = 2s \quad \text{and} \quad [\text{B}^{2-}] = s ]

giving

[ K_{sp} = (2s)^2 \times s = 4s^3 ]

In general, if the dissolution reaction is

[ \text{A}_a\text{B}_b(s) \rightleftharpoons a\text{A}^{b+}(aq) + b\text{B}^{a-}(aq) ]

then

[ K_{sp} = (as)^a \times (bs)^b ]

and s can be isolated by taking the appropriate root Less friction, more output..

Step‑by‑Step Procedure to Calculate Molar Solubility

1. Write the balanced dissolution equation for the solid, including the correct stoichiometric coefficients.
2. Express the equilibrium concentrations of each ion in terms of the unknown molar solubility s (multiply s by the coefficient from the equation).
3. Formulate the Ksp expression using those concentrations.
4. Substitute the known Ksp value (usually given in tables or problems) into the expression.
5. Solve for s by isolating it algebraically; this may involve taking a square root, cube root, or higher‑order root depending on the exponents.
6. Check units – the result should be in mol L⁻¹. If the problem asks for solubility in grams per liter, multiply s by the molar mass of the solid.
7. Consider any complicating factors (common ion effect, pH, complexation) and adjust the calculation accordingly (see the “Factors Affecting Molar Solubility” section).

Example Problems

Example 1: Simple 1:1 Salt (Silver Chloride)

Silver chloride dissociates as

[ \text{AgCl(s)} \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) ]

Given (K_{sp} = 1.8 \times 10^{-10}) at 25 °C And that's really what it comes down to..

1. Let the molar solubility be s. Then ([\text{Ag}^+] = s) and ([\text{Cl}^-] = s).
2. (K_{sp} = s \times s = s^2).
3. (s^2 = 1.8 \times 10^{-10}).
4. (s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}) mol L⁻¹.

Thus, the molar solubility of AgCl is (1.3 \times 10^{-5}) M. In grams per liter:

[ 1.On the flip side, 34 \times 10^{-5},\text{mol L}^{-1} \times 143. 32,\text{g mol}^{-1} \approx 1 Less friction, more output..

Example 2: Salt with Different Stoichiometry (Lead(II) Iodide)

Lead(II) iodide dissolves as

[ \text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) ]

(K_{sp} = 7.1 \times 10^{-9}) Most people skip this — try not to..

1. Let solubility be s. Then ([\text{Pb}^{2+}] = s) and ([\text{I}^-] = 2s).
2. (K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 = s \times (2s)^2 = 4s^3).
3. (4s^3 = 7.1 \times 10^{-9}).
4. (s^3 = \frac{7.1 \times 10^{-9}}{4} = 1.775 \times 10^{-9}).
5. (s = \sqrt[3]{1.775 \times 10^{-9}} = 1.22 \times 10^{-3}) mol L⁻¹.

Molar solubility of PbI₂ ≈ (1.2 \times 10^{-3}) M. In g L⁻¹:

[ 1.22 \times 10^{-