# Solubility Equilibria

Chemistry I · Energy, Equilibrium and Electrochemistry · https://tryals.app/learn/chemistry-i/solubility-equilibria

## How Insoluble Is Insoluble?

No salt is completely insoluble. A sparingly soluble solid in contact with its saturated solution is an equilibrium, described by the **solubility product**. For $\mathrm{A_xB_y(s) \rightleftharpoons xA^{y+} + yB^{x-}}$:

$$K_{sp} = [\mathrm{A^{y+}}]^x[\mathrm{B^{x-}}]^y$$

The solid itself does not appear, being a pure phase. Converting $K_{sp}$ into **molar solubility** $s$ requires care with the stoichiometry:

| Salt type | Relation | Example |
|---|---|---|
| AB | $K_{sp} = s^2$ | AgCl |
| $\mathrm{AB_2}$ | $K_{sp} = 4s^3$ | $\mathrm{CaF_2}$ |
| $\mathrm{A_2B}$ | $K_{sp} = 4s^3$ | $\mathrm{Ag_2CrO_4}$ |

The factor of 4 arises because two ions form per formula unit, so that ion's concentration is $2s$ and it is squared. Comparing $K_{sp}$ values across different salt types is therefore meaningless, silver chromate has a smaller $K_{sp}$ than silver chloride yet is *more* soluble.

The **common ion effect** follows from Le Chatelier: adding an ion the solid already contains pushes the equilibrium back toward the solid, so solubility falls. Silver chloride is markedly less soluble in sodium chloride solution than in pure water.

Whether a precipitate forms is decided by comparing the **ion product** $Q$ with $K_{sp}$: if $Q > K_{sp}$ the solution is supersaturated and solid appears; if $Q < K_{sp}$ any solid present dissolves. **Selective precipitation** exploits the gap between two salts' $K_{sp}$ values, adding a reagent slowly so the less soluble one drops out first.

> **Common pitfall:** comparing $K_{sp}$ values of salts with different formulas. $K_{sp}$ has different units and different exponents for AB and $\mathrm{AB_2}$ salts, so only solubilities computed from them are comparable, a smaller $K_{sp}$ does not reliably mean a less soluble salt.

## Practice questions

8 of this lesson's 12 practice questions, with answers. The full set is in the app.

### 1. Complete the explanation of why two solubility products cannot be compared directly.

**Answer:** For a 1:1 salt the solubility product equals **s squared**, but for an $\mathrm{A_2B}$ salt it equals **4s cubed**. Because the two are different functions of the solubility, silver chromate can have the **smaller** constant while being the **more** soluble of the two.

**Why:** For AgCl, $K_{sp} = s^2$; for $\mathrm{Ag_2CrO_4}$, $K_{sp} = 4s^3$. Cubing a small solubility pushes the constant far lower for the same amount dissolved, so only solubilities computed from the constants are comparable, never the constants themselves.

Page: https://tryals.app/practice/chemistry-i/solubility-equilibria/complete-the-explanation-of-why-two-solubility-products-cannot-be

### 2. A solution chemist claims that the solubility product constant reflects the equilibrium position of dissolution, yet it cannot serve as a universal scale for solubility. What underlies this distinction?

A. Precipitation shifts the equilibrium whenever any foreign electrolyte is present
B. Solids are excluded from the expression because their activity remains zero
C. Ionic exponents depend upon stoichiometry rather than total dissolved moles
D. Calculations of molar solubility are valid only in entirely neutral solutions

**Answer:** C. Ionic exponents depend upon stoichiometry rather than total dissolved moles

**Why:** Equilibrium constants scale with stoichiometric powers rather than bare moles dissolved. Assuming lower constants always mean lower solubility conflates thermodynamic constants with actual dissolved concentrations across different lattice stoichiometries.

Page: https://tryals.app/practice/chemistry-i/solubility-equilibria/a-solution-chemist-claims-that-the-solubility-product-constant

### 3. A salt of formula $\mathrm{AB_2}$ has a molar solubility of $1.0 \times 10^{-2}$ M. Compute its solubility product, in units of $10^{-6}$.

**Answer:** 4 (within ±0.2)

**Why:** $K_{sp} = [A][B]^2 = s(2s)^2 = 4s^3 = 4 \times (10^{-2})^3 = 4 \times 10^{-6}$. Forgetting either the doubling or the squaring is the classic error, and each one changes the answer by a factor of four.

Page: https://tryals.app/practice/chemistry-i/solubility-equilibria/a-salt-of-formula-ab2-has-a-molar-solubility-of-1-0-10-m-compute

### 4. Sort each condition by what happens to a sparingly soluble salt in the solution.

**Answer:**

- Precipitate forms: Ion product greater than the solubility product, A common ion is added to a saturated solution
- Solid dissolves: Ion product less than the solubility product
- System already at equilibrium: Ion product equal to the solubility product

**Why:** The comparison of $Q$ with $K_{sp}$ decides everything: above it the solution is supersaturated and solid appears, below it any solid dissolves, and equality is saturation. Adding a common ion raises $Q$ and forces precipitation.

Page: https://tryals.app/practice/chemistry-i/solubility-equilibria/sort-each-condition-by-what-happens-to-a-sparingly-soluble-salt-in

### 5. Adding sodium chloride to a saturated solution of silver chloride increases the amount of silver chloride that dissolves.

**Answer:** False

**Why:** False, this is the common ion effect working in the opposite direction. Added chloride drives the equilibrium back toward solid silver chloride, so *less* dissolves, not more.

Page: https://tryals.app/practice/chemistry-i/solubility-equilibria/adding-sodium-chloride-to-a-saturated-solution-of-silver-chloride

### 6. A precipitating reagent is added slowly to a solution containing two metal ions with very different solubility products. Order what happens.

**Answer:**

1. The ion product of the less soluble salt reaches its solubility product first
2. The less soluble salt begins to precipitate
3. The more soluble salt reaches its own solubility product later
4. Both salts precipitate together and separation is lost

**Why:** Adding the reagent slowly lets the less soluble salt saturate and precipitate on its own. Only once enough reagent has been added does the second salt reach its own constant, at which point the two co-precipitate and separation is lost, so the reagent is stopped before that.

Page: https://tryals.app/practice/chemistry-i/solubility-equilibria/a-precipitating-reagent-is-added-slowly-to-a-solution-containing-two

### 7. Match each salt formula type to its relation between solubility product and molar solubility.

**Answer:**

- AB, such as silver chloride → $K_{sp} = s^2$
- $\mathrm{AB_2}$, such as calcium fluoride → $K_{sp} = 4s^3$
- $\mathrm{A_3B}$, such as silver phosphate → $K_{sp} = 27s^4$
- Any pure solid phase → Omitted from the expression entirely

**Why:** Each ion appears at its stoichiometric multiple of $s$ and raised to its coefficient, which is where the numerical prefactors 4 and 27 come from. Pure solids never appear, since their concentration cannot vary.

Page: https://tryals.app/practice/chemistry-i/solubility-equilibria/match-each-salt-formula-type-to-its-relation-between-solubility

### 8. A 1:1 salt has $K_{sp} = 9.0 \times 10^{-12}$. Compute its molar solubility in pure water, in units of $10^{-6}$ M.

**Answer:** 3 (within ±0.15)

**Why:** $s = \sqrt{9.0 \times 10^{-12}} = 3.0 \times 10^{-6}$ M. Note how a constant of order $10^{-12}$ corresponds to a solubility of order $10^{-6}$, the square root halves the exponent.

Page: https://tryals.app/practice/chemistry-i/solubility-equilibria/a-1-1-salt-has-ksp-9-0-10-compute-its-molar-solubility-in