Removing an ion makes more solid dissolve
In Topic 7.12 you added one of a salt’s ions and the salt became less soluble. This topic looks at the opposite stress: something in the solution removes one of the salt’s ions. The tool is the same, Le Châtelier’s principle.
Take calcium fluoride in water, with some undissolved solid present:
CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq) K_sp = [Ca²⁺][F⁻]²
F⁻ is the conjugate base of HF, a weak acid. So F⁻ is a weak base. If you add a strong acid, H₃O⁺ reacts with F⁻:
F⁻(aq) + H₃O⁺(aq) → HF(aq) + H₂O(l)
Follow the stress step by step:
- H₃O⁺ turns F⁻ into HF, so [F⁻] falls.
- Q = [Ca²⁺][F⁻]² is now less than K_sp.
- The dissolving reaction runs forward to replace the lost F⁻: more CaF₂ dissolves.
- More Ca²⁺ ends up in solution. The salt is more soluble in acid than in pure water.
Adding the two equations (after doubling the second) gives the overall change:
CaF₂(s) + 2 H₃O⁺(aq) → Ca²⁺(aq) + 2 HF(aq) + 2 H₂O(l)
K_sp itself has not changed. It is still an equilibrium constant that depends only on temperature. What changes is how much solid has to dissolve before the ion product reaches K_sp again.
Which salts are sensitive to pH?
Look at each ion in the salt and ask: can this ion gain or lose a proton? The solubility depends on pH when one of the ions is a weak base, a weak acid or the hydroxide ion itself.
| The salt contains… | Examples | What removes the ion | Effect on solubility |
|---|---|---|---|
| An anion that is a weak base (conjugate base of a weak acid) | CaF₂, CaCO₃, Ca₃(PO₄)₂, CaC₂O₄ | H₃O⁺ turns the anion into its conjugate acid | Increases as pH falls |
| Hydroxide ion | Mg(OH)₂, Fe(OH)₃ | H₃O⁺ turns OH⁻ into water | Increases as pH falls; decreases as pH rises (OH⁻ is a common ion) |
| A cation that is a weak acid, such as BH⁺, the conjugate acid of a weak base B | the fictional salt (BH)X | OH⁻ turns BH⁺ into B | Increases as pH rises |
| Only ions that are neither acids nor bases in water | AgCl, AgBr, AgI | nothing: H₃O⁺ and OH⁻ do not react with them | No change with pH (in this course’s model) |
Cl⁻, Br⁻ and I⁻ are conjugate bases of strong acids. They have almost no tendency to take a proton, so acid cannot remove them.
How much does the pH have to change? Use the idea from Topic 8.7. When the pH is above the pK_a of the weak acid HA, the base form A⁻ predominates. Little A⁻ is removed, and the solubility is close to its value in pure water. When the pH drops below the pK_a, the acid form HA predominates. Most of the dissolved A⁻ is turned into HA, so much more solid must dissolve. For a salt with a weakly acidic cation BH⁺, the same reasoning works the other way round: the solubility rises once the pH climbs above the pK_a of BH⁺.
Picturing the effect
Two features are worth noticing. First, the hydroxide (line 4) responds to pH everywhere, because [OH⁻] is fixed directly by the pH. Second, a salt whose ion is a very weak base, such as the salt in line 1, only responds once the solution is acidic enough to protonate that ion. You only need the direction of each change and the reason for it.
Worked example 1: predicting the effect of acid
Question. Nitric acid is added to separate beakers, each containing water and some undissolved solid: (a) calcium carbonate, CaCO₃; (b) silver bromide, AgBr; (c) magnesium hydroxide, Mg(OH)₂; (d) barium fluoride, BaF₂. For each, predict whether the solubility increases, decreases or stays the same, and justify your answer. Then (e) predict what adding NaOH does to (c).
Method. Write the dissolving equation, then ask whether H₃O⁺ reacts with either ion.
(a) CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃²⁻(aq). Carbonate is the conjugate base of the weak acid HCO₃⁻, so it is a base. H₃O⁺ converts it to HCO₃⁻ and then H₂CO₃, which breaks down to CO₂ and water. [CO₃²⁻] falls, Q < K_sp, so more CaCO₃ dissolves: increases. Overall, with excess acid:
CaCO₃(s) + 2 H₃O⁺(aq) → Ca²⁺(aq) + CO₂(g) + 3 H₂O(l)
The CO₂ bubbles out of the solution, so the carbonate cannot come back. That is why a limestone chip fizzes and disappears in dilute acid.
(b) AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq). Br⁻ is the conjugate base of HBr, a strong acid, so it does not accept a proton. Ag⁺ does not react with H₃O⁺. Q is unchanged: stays the same. (NO₃⁻ from the nitric acid is not a common ion either.)
(c) Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2 OH⁻(aq). H₃O⁺ + OH⁻ → 2 H₂O removes hydroxide, Q < K_sp: increases. Overall: Mg(OH)₂(s) + 2 H₃O⁺(aq) → Mg²⁺(aq) + 4 H₂O(l). This is how a magnesium hydroxide antacid neutralises stomach acid.
(d) BaF₂(s) ⇌ Ba²⁺(aq) + 2 F⁻(aq). F⁻ is a weak base (HF is a weak acid). H₃O⁺ converts F⁻ to HF, Q < K_sp: increases.
(e) NaOH and Mg(OH)₂. OH⁻ is a common ion. Q rises above K_sp and Mg(OH)₂ precipitates until Q = K_sp again: the solubility decreases.
Check. The only salt that does not respond is the one whose anion comes from a strong acid. That fits the rule in the table.
Worked example 2: reading solubility data from the lab
Question. A student studies three fictional white solids, P, Q and R. She stirs 0.800 g of each solid with 100.0 mL of a buffer until no more dissolves, then filters, dries and weighs the solid left over. The balance reads to ±0.001 g. None of the buffers contains an ion that is also in P, Q or R.
| Solid | Mass recovered at pH 3 (g) | at pH 7 (g) | at pH 11 (g) |
|---|---|---|---|
| P | 0.796 | 0.795 | 0.796 |
| Q | 0.740 | 0.794 | 0.794 |
| R | 0.782 | 0.782 | 0.619 |
(a) Calculate the mass of each solid that dissolved at each pH. (b) Classify each solid as having a basic anion, a weakly acidic cation, or neither. (c) Why did the student need buffers, rather than plain acid or base, and why must the buffer not contain a common ion?
(a) Mass dissolved = 0.800 g − mass recovered.
| Solid | pH 3 | pH 7 | pH 11 |
|---|---|---|---|
| P | 0.004 g | 0.005 g | 0.004 g |
| Q | 0.060 g | 0.006 g | 0.006 g |
| R | 0.018 g | 0.018 g | 0.181 g |
(b) P: the three values differ by only 0.001 g, which is the precision of the balance. There is no real change with pH, so neither ion of P is acidic or basic. Q: about ten times as much dissolves at pH 3 as at pH 7, and pH 7 and pH 11 give the same result. Acid removes one of Q’s ions, so Q has a basic anion. Because nothing changes between pH 7 and 11, the anion is a base whose conjugate acid has a pK_a below 7. R: the mass dissolved is unchanged from pH 3 to pH 7 but about ten times larger at pH 11. Base removes one of R’s ions, so R has a weakly acidic cation.
(c) Dissolving Q uses up H₃O⁺, and dissolving R uses up OH⁻. In plain acid or base, the pH would drift as the solid dissolved. A buffer with enough capacity (Topic 8.10) keeps the pH nearly constant, so each result belongs to one pH value. A common ion from the buffer would lower the solubility (Topic 7.12) and hide the pH effect.
Lesson. When you read data, compare differences with the precision of the measurement before you claim a trend.
Worked example 3: acid and tooth enamel
Question. The mineral in tooth enamel is mainly hydroxyapatite, Ca₅(PO₄)₃OH. Bacteria in the mouth turn sugars into acids. Explain why this acid can dissolve enamel.
Answer. In water, a little hydroxyapatite dissolves:
Ca₅(PO₄)₃OH(s) ⇌ 5 Ca²⁺(aq) + 3 PO₄³⁻(aq) + OH⁻(aq)
Two of its ions are bases. H₃O⁺ converts PO₄³⁻ to HPO₄²⁻ (and further), and converts OH⁻ to water. Both changes lower Q below K_sp, so the equilibrium shifts to the right and more enamel dissolves. The lower the pH near the tooth surface, the more enamel can dissolve before equilibrium is reached.
Extension. Fluoride treatments help form fluorapatite, which resists acid attack better than hydroxyapatite. One part of the reason is that F⁻ is a much weaker base than OH⁻, so acid removes it far less readily.
What you will not be asked to do
The course asks for qualitative reasoning only. You need to predict the direction of the change, write the reaction that removes an ion, and explain the shift with Le Châtelier’s principle. Calculating solubility at a given pH is not assessed. You may still meet K_sp calculations (Topic 7.11) and pH calculations (Topics 8.2 and 8.3) on their own.
Common misconceptions
- “Acid makes every salt more soluble.” Only salts with a basic anion or hydroxide. AgCl, AgBr and AgI do not respond to acid.
- “The pH changes K_sp.” K_sp depends only on temperature. The pH changes the solubility by removing ions, not the constant.
- “H₃O⁺ reacts with the metal cation.” In these examples the acid reacts with the anion: CO₃²⁻, F⁻, PO₄³⁻ or OH⁻.
- “To dissolve a chloride, add HCl.” The Cl⁻ in hydrochloric acid is a common ion. It can only lower the solubility of a chloride salt.
- “Adding base always lowers solubility.” It lowers the solubility of a hydroxide (common ion), but it raises the solubility of a salt whose cation is a weak acid.
- “Any change in the data is a real trend.” A difference no larger than the balance’s precision is not evidence of a pH effect (Worked example 2, solid P).
Where this leads
This topic ends Unit 8 and joins it to Unit 7: acid-base reactions are another way to apply a stress to a solubility equilibrium. Next, Topic 9.1 begins thermodynamics; in Topic 9.6 you will see why dissolving is favourable for some salts and not others. Try the practice questions now, then use the revision notes and the checklist to consolidate.
Quick questions
Why does silver chloride not dissolve in nitric acid?
Cl⁻ is the conjugate base of a strong acid, so it has almost no tendency to accept a proton. H₃O⁺ cannot remove it from solution, Q stays equal to K_sp, and the solubility does not change.
Does a hydroxide dissolve less in base?
Yes. Added OH⁻ is a common ion for a salt such as Mg(OH)₂, so the equilibrium shifts towards the solid. This is the common-ion effect from Topic 7.12.