⏱️General Chemistry II Unit 5 Review

The solubility product constant ( $K_{sp}$ ) describes the equilibrium between a sparingly soluble ionic compound and its dissolved ions in water. It lets you predict how much of a salt will dissolve, compare solubilities across compounds, and understand how conditions like pH shift that equilibrium.

Definition of Ksp

When a sparingly soluble salt is added to water, it dissolves until the solution becomes saturated. At that point, the rate of dissolving equals the rate of precipitation, and a dynamic equilibrium exists between the solid and its ions. The $K_{sp}$ is the equilibrium constant for this dissolution reaction.

For a generic salt $A_aB_b$ :

$A_aB_b(s) \rightleftharpoons aA^{n+}(aq) + bB^{m-}(aq)$

$K_{sp} = [A^{n+}]^a[B^{m-}]^b$

A few things to keep in mind:

The solid does not appear in the $K_{sp}$ expression (just like any equilibrium involving a pure solid — its activity is 1).
$K_{sp}$ is temperature-dependent. By strict thermodynamic convention it's dimensionless (since it's defined using activities), but you'll often see it treated as having no units in general chemistry courses.
A smaller $K_{sp}$ means lower solubility when comparing salts of the same stoichiometric type. For example, $AgCl$ ( $K_{sp} = 1.8 \times 10^{-10}$ ) is less soluble than $CuCl$ ( $K_{sp} = 1.7 \times 10^{-7}$ ), and both are 1:1 salts, so the direct comparison works.

Definition of solubility product constant, Solubility Product | Introduction to Chemistry

Calculations with Ksp

The two most common calculation types are (1) finding $K_{sp}$ from molar solubility and (2) finding molar solubility from $K_{sp}$ . Molar solubility ( $s$ ) is the number of moles of solute that dissolve per liter of saturated solution.

Finding Ksp from Molar Solubility

Write the balanced dissolution equation.
Define each ion concentration in terms of $s$ using stoichiometric ratios.
Substitute into the $K_{sp}$ expression and simplify.

Example: $CaF_2$ has a molar solubility of $3.3 \times 10^{-4}$ M.

$CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$

$[Ca^{2+}] = s$ , and $[F^-] = 2s$

$K_{sp} = (s)(2s)^2 = 4s^3 = 4(3.3 \times 10^{-4})^3 = 1.4 \times 10^{-10}$

Finding Molar Solubility from Ksp

Write the balanced dissolution equation.
Express each ion concentration in terms of $s$ .
Plug into the $K_{sp}$ expression and solve for $s$ .

Example: $Ag_2CrO_4$ has $K_{sp} = 1.1 \times 10^{-12}$ .

$Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$

$[Ag^+] = 2s$ , and $[CrO_4^{2-}] = s$

$K_{sp} = (2s)^2(s) = 4s^3$

$s = \left(\frac{1.1 \times 10^{-12}}{4}\right)^{1/3} = 6.5 \times 10^{-5} \text{ M}$

A common mistake here is forgetting the coefficient in front of $s$ . If two moles of an ion are produced per formula unit, you must write $2s$ and then square it, giving you $4s^2$ in the expression. Missing that factor of 4 will throw off your answer significantly.

Definition of solubility product constant, Basics of solubility and Solubility Products

Relative Solubility from Ksp

You can compare the solubilities of different compounds using $K_{sp}$ , but there's an important caveat: direct comparison of $K_{sp}$ values only works reliably when the compounds produce the same total number of ions (the same stoichiometric type).

For salts of the same type (e.g., all 1:1 like $AgCl$ vs. $AgBr$ , or all 1:2 like $CaF_2$ vs. $BaF_2$ ), a larger $K_{sp}$ means greater molar solubility. You can rank them directly.
For salts of different types (e.g., $AgCl$ which is 1:1 vs. $Ag_2CrO_4$ which is 2:1), the $K_{sp}$ expressions have different algebraic forms ( $s^2$ vs. $4s^3$ ). In these cases, you need to calculate the actual molar solubility for each compound before comparing.

This is a point that shows up frequently on exams. If a question asks you to rank solubilities and the salts have different formulas, don't just look at the $K_{sp}$ values. Solve for $s$ for each one.

pH Effects on Salt Solubility

The solubility of certain salts changes with pH. This happens whenever one of the dissolved ions can react with $H^+$ or $OH^-$ , removing that ion from solution and shifting the dissolution equilibrium forward (Le Chatelier's principle).

Salts with basic anions become more soluble in acid. If the anion is the conjugate base of a weak acid (e.g., $CO_3^{2-}$ , $PO_4^{3-}$ , $F^-$ , $S^{2-}$ , $OH^-$ ), adding $H^+$ protonates that anion and removes it from solution. The equilibrium shifts toward dissolving more solid.

For $CaCO_3$ : In acidic solution, $CO_3^{2-}$ reacts with $H^+$ to form $HCO_3^-$ (and eventually $CO_2$ gas), pulling the equilibrium to the right and increasing solubility. This is why calcium carbonate dissolves in acid rain but not in pure water.

The key test is: Is the anion the conjugate base of a weak acid? If yes, the salt's solubility is pH-dependent. If the anion comes from a strong acid, it won't react with $H^+$ in any meaningful way.

Salts with anions from strong acids are unaffected by pH. Anions like $Cl^-$ , $Br^-$ , $I^-$ , and $NO_3^-$ don't react appreciably with $H^+$ , so changing pH doesn't shift the equilibrium. The solubility of $AgCl$ , for instance, is essentially the same whether the solution is acidic or neutral. ( $SO_4^{2-}$ is a borderline case since $HSO_4^-$ is a weak acid with $K_a \approx 0.012$ , but for most Gen Chem purposes it's treated as fully dissociated and pH-independent.)

Note: $AgCl$ solubility does increase in the presence of $NH_3$ , but that's a complex ion formation effect (covered in the next section on complex ions), not a simple pH effect. The $NH_3$ acts as a Lewis base that binds $Ag^+$ to form $[Ag(NH_3)_2]^+$ , removing $Ag^+$ from solution and shifting the equilibrium.

⏱️General Chemistry II Unit 5 Review