Premed · Premed · General Chemistry 2
Lecture 9: Solubility Equilibria and Ksp
General Chemistry II
Learning Objectives
By the end of this lecture, students will be able to:
- Write solubility product expressions for sparingly soluble salts
- Calculate Ksp from solubility data and vice versa
- Predict whether a precipitate will form using the ion product Q
- Explain and calculate the common-ion effect on solubility
- Describe the effect of pH on solubility
- Apply selective precipitation to separate ions
Lecture Content
I. The Solubility Product Constant (Ksp)
Sparingly soluble ionic compounds establish an equilibrium between the undissolved solid and the ions in solution. For a generic salt M_aX_b(s) <=> aM^n+(aq) + bX^m-(aq), the equilibrium constant expression is Ksp = [M^n+]^a * [X^m-]^b. The solid does not appear in the expression because its activity is defined as 1. The solubility product constant Ksp is a special case of the general equilibrium constant Keq, applied specifically to dissolution equilibria. Like all equilibrium constants, Ksp depends only on temperature. A smaller Ksp generally indicates lower solubility, though direct comparisons are only valid for salts with the same ion ratio.
Representative examples include AgCl(s) <=> Ag+(aq) + Cl-(aq) with Ksp = [Ag+][Cl-] = 1.8 x 10^-10, PbI2(s) <=> Pb^2+(aq) + 2I-(aq) with Ksp = [Pb^2+][I-]^2 = 9.8 x 10^-9, and Ca3(PO4)2(s) <=> 3Ca^2+(aq) + 2PO4^3-(aq) with Ksp = [Ca^2+]^3[PO4^3-]^2.
II. Calculating Ksp from Solubility
The molar solubility, s, is the number of moles of solute that dissolve per liter of saturated solution. To calculate Ksp from molar solubility, write the dissolution equilibrium, use stoichiometry to express each ion concentration in terms of s, and substitute into the Ksp expression.
For AgCl, with a molar solubility of 1.3 x 10^-5 M, both [Ag+] and [Cl-] equal s, so Ksp = s^2 = (1.3 x 10^-5)^2 = 1.7 x 10^-10. For PbI2, with a molar solubility of 1.5 x 10^-3 M, [Pb^2+] = s and [I-] = 2s, giving Ksp = s * (2s)^2 = 4s^3 = 4(1.5 x 10^-3)^3 = 1.4 x 10^-8.
III. Calculating Solubility from Ksp
The reverse calculation sets up the dissolution equilibrium, expresses the ion concentrations in terms of s, and solves for s. An important caveat is that you can only compare Ksp values directly to rank solubility when the salts have the same stoichiometric ratio of ions (for example, all 1:1 or all 1:2 salts). For salts with different ion ratios, you must calculate the actual molar solubilities and compare those. Converting between units is straightforward: multiplying the molar solubility (mol/L) by the molar mass gives the solubility in g/L.
<image>A comparison table showing Ksp and molar solubility calculations for three salts with different stoichiometries. Column headers: Salt, Equilibrium, Ksp Expression, Ksp Value, Molar Solubility (s). Row 1: AgCl (1:1), Ksp = s^2, Ksp = 1.8 x 10^-10, s = 1.3 x 10^-5 M. Row 2: PbF2 (1:2), Ksp = 4s^3, Ksp = 3.3 x 10^-8, s = 2.0 x 10^-3 M. Row 3: Ca3(PO4)2 (3:2), Ksp = 108s^5, Ksp = 2.1 x 10^-33, s = 1.1 x 10^-7 M. A note at the bottom emphasizes that PbF2 has a larger Ksp than AgCl but their relative solubilities require calculation of s.</image>
IV. Predicting Precipitation: Q vs. Ksp
The ion product Q is calculated using the same expression as Ksp, but with the current (non-equilibrium) concentrations of the ions. Comparing Q to Ksp reveals the state of the solution. If Q < Ksp, the solution is unsaturated and no precipitate forms; more solid could still dissolve. If Q = Ksp, the solution is exactly saturated and at equilibrium. If Q > Ksp, the solution is supersaturated and a precipitate will form, continuing until Q decreases to equal Ksp. When mixing solutions, always remember to account for dilution by recalculating concentrations using the new total volume: new concentration = moles of ion / total volume after mixing.
V. The Common-Ion Effect on Solubility
The solubility of a sparingly soluble salt decreases when a common ion is already present in solution. This is a direct application of Le Chatelier's principle: adding an ion that appears in the dissolution equilibrium shifts the equilibrium to the left, reducing solubility.
For example, the solubility of AgCl in 0.10 M NaCl is dramatically lower than in pure water. The NaCl provides Cl- at 0.10 M as a common ion. Setting up the Ksp expression, Ksp = [Ag+][Cl-] = (s)(0.10 + s), and approximating 0.10 + s as 0.10, gives s = 1.8 x 10^-10 / 0.10 = 1.8 x 10^-9 M. This represents a decrease of nearly four orders of magnitude compared to the solubility in pure water (1.3 x 10^-5 M). The common-ion effect is important in controlling precipitation reactions and plays significant roles in biological systems.
VI. Effect of pH on Solubility
Salts whose anion is the conjugate base of a weak acid become more soluble in acidic solution. The H+ ions react with the basic anion, removing it from solution and shifting the dissolution equilibrium to the right. For CaF2, the F- ions react with H+ to form the weak acid HF, thereby increasing the solubility of CaF2. For CaCO3, the effect is even more dramatic: CO3^2- reacts with H+ to form H2CO3, which decomposes to H2O and CO2 gas, pulling the equilibrium strongly to the right.
Salts whose anions come from strong acids, such as Cl- and NO3-, are not affected by pH because these anions are too weak as bases to react with H+. Metal hydroxides, M(OH)n, become more soluble in acidic solution because the hydroxide ions are neutralized: OH- + H+ -> H2O. Practical consequences of pH-dependent solubility include the dissolution of marble (CaCO3) buildings by acid rain and the action of stomach acid on calcium carbonate antacid tablets.
<image>A diagram showing the effect of pH on the solubility of CaCO3. Panel A: In neutral water, CaCO3(s) is at equilibrium with Ca^2+ and CO3^2- (low solubility). Panel B: In acidic solution, H+ ions react with CO3^2- to form HCO3- and then H2CO3, which decomposes to H2O and CO2 (shown as bubbles). The removal of CO3^2- shifts the dissolution equilibrium to the right, dramatically increasing solubility. Arrows and equilibrium expressions illustrate the coupled equilibria.</image>
VII. Selective Precipitation
Selective precipitation is a technique for separating ions in solution by adding a reagent that precipitates one ion while leaving another in solution. The procedure involves calculating the concentration of the precipitating anion needed to just begin precipitating each cation. The ion that requires the lower concentration of the precipitating anion precipitates first. By slowly adding the reagent, you can selectively precipitate one ion while keeping the other in solution.
For example, Ag+ and Pb^2+ can be separated using Cl-. Because AgCl has a much smaller Ksp (1.8 x 10^-10) than PbCl2 (1.7 x 10^-5), AgCl precipitates first at a much lower Cl- concentration. This principle underlies qualitative analysis in the laboratory, where systematic selective precipitation is used to identify unknown ions in solution.
VIII. Fractional Precipitation and Completeness
An ion is considered "completely" precipitated when 99.9% or more has been removed from solution. To verify the feasibility of a separation, calculate the concentration of the precipitating ion needed to reduce the target ion to 0.1% of its original concentration, and then confirm that this concentration does not exceed the threshold at which the second ion would begin to precipitate.
