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Buffer Solutions and Solubility Product

A buffer is a solution that keeps its pH nearly the same when a little acid or base is added. An acidic buffer is a weak acid with its salt (CH₃COOH + CH₃COONa); a basic buffer is a weak base with its salt (NH₄OH + NH₄Cl). Its pH is given by the Henderson equation: pH = pKa + log([salt]/[acid]). A sparingly soluble salt in its saturated solution sets up an equilibrium with its ions; the product of ion concentrations (each raised to its coefficient) is the solubility product Ksp. For AB, Ksp = s²; for AB₂ or A₂B, Ksp = 4s³. If the ionic product Q exceeds Ksp, a precipitate forms. A common ion lowers solubility.

🎬 Step-by-step story

  1. Two beakers get 1 mmol of HCl each. Pure water's pH crashes from 7 to 2. In the buffer, each H⁺ is caught by one acetate ion (purple) and turned into acetic acid (green). The buffer's pH barely moves.
  2. A buffer's pH depends on the ratio of salt to acid: pH = pKa + log([salt]/[acid]). Equal amounts give pH = pKa = 4.74. Slide the ratio and watch the purple and green counts.
  3. Now a crystal of AgCl sits in water. Only a tiny amount dissolves. Ions keep leaving the crystal and joining it back. The product [Ag⁺][Cl⁻] stays fixed: this is Ksp.
  4. From Ksp we get the solubility s. For AgCl (1 : 1), Ksp = s². For CaF₂ (1 : 2), there are two F⁻ for every Ca²⁺, so Ksp = s × (2s)² = 4s³.
  5. Add NaCl. Now there is extra Cl⁻, so [Ag⁺][Cl⁻] becomes bigger than Ksp. Ag⁺ is pushed back into the solid as a precipitate. A common ion makes the salt less soluble.
  6. Free play: add acid or base to the buffer, or pick a salt and add a common ion.

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Why doesn't the added H⁺ lower the buffer's pH?

Acetate ions catch the H⁺ and turn into acetic acid, which hardly ionises. In the 3D each drop turns one purple ion green.

Why is pH = pKa the best buffer?

Equal acid and salt can absorb equal amounts of added base or acid. The slider shows the counts are equal at ratio 1.

If some AgCl dissolves, why doesn't all of it?

Ions also return to the crystal. When leaving and returning are equal, the solution is saturated and [Ag⁺][Cl⁻] = Ksp.

Why is Ksp = 4s³ for CaF₂, not s²?

Each CaF₂ gives one Ca²⁺ and two F⁻, so [F⁻] = 2s and Ksp = s(2s)². The 3D shows twice as many F⁻ ions.

Why does NaCl make AgCl less soluble?

Extra Cl⁻ makes Q bigger than Ksp, so Ag⁺ and Cl⁻ join back into the solid until Q = Ksp again. Watch the crystal grow in the 3D.

Does adding water to a buffer change its pH?

Hardly. Both salt and acid are diluted equally, so their ratio stays the same.

Buffer solutions: what they are and how they work

A buffer solution resists a change in pH when a little acid or base is added, or when it is diluted.

How it works (acetate buffer): the salt gives lots of CH₃COO⁻, and the weak acid stays mostly as CH₃COOH (the common ion stops it ionising).

Only the ratio [salt]/[acid] changes a little, so pH changes very little. The buffer works best when this ratio is between 0.1 and 10 (pH = pKa ± 1). Its buffer capacity is larger when both are more concentrated.

Try it: at step 1 press “+ 1 mmol HCl” again and again. Water's pH drops at once; the buffer's pH falls slowly until the purple A⁻ runs out.

Henderson–Hasselbalch equation (derivation)

Start from the weak acid: Ka = [H⁺][A⁻]/[HA].

Rearrange: [H⁺] = Ka × [HA]/[A⁻]. Take −log of both sides:

−log[H⁺] = −log Ka − log([HA]/[A⁻]) → pH = pKa + log([salt]/[acid]).

Here [A⁻] ≈ [salt] (the salt is fully ionised) and [HA] ≈ [acid] (the acid hardly ionises). For a basic buffer: pOH = pKb + log([salt]/[base]), then pH = 14 − pOH.

When [salt] = [acid], log 1 = 0, so pH = pKa. This is the best buffering point.

Solubility product of sparingly soluble salts

Salts like AgCl, BaSO₄ and CaF₂ dissolve only a very little. In a saturated solution the solid is in equilibrium with its ions: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).

The solid is left out, so Ksp = [Ag⁺][Cl⁻]. In general, for AₓBᵧ(s) ⇌ xAʸ⁺ + yBˣ⁻: Ksp = [Aʸ⁺]ˣ[Bˣ⁻]ʸ.

Solubility s (mol/L) and Ksp:

Compare solubility only through s, not Ksp, when the salts have different formula types.

Ionic product, precipitation and the common ion effect

The ionic product Q is the same expression but with the present concentrations.

Common ion effect: adding an ion already present (Cl⁻ from NaCl to AgCl) pushes the equilibrium back, so solubility falls. In 0.1 M NaCl, AgCl's solubility is s = Ksp/0.1 = 1.8 × 10⁻⁹ M, about 7 000 times less than in pure water.

This is used in salt analysis (group separation using H₂S in acid or in NH₄OH/NH₄Cl) and to purify common salt.

Try it: at step 5 press “add NaCl” and see ions return to the crystal.

Exam corner

Expect numericals on buffer pH, ratio needed for a target pH, Ksp from s, s from Ksp for 1:1 and 1:2 salts, whether a precipitate forms (Q vs Ksp), and solubility in the presence of a common ion.

Key formulas and definitions

Worked examples

1. Find the pH of a buffer with 0.1 M CH₃COOH and 0.1 M CH₃COONa (pKa = 4.74).

Ratio = 1, log 1 = 0, so pH = pKa = 4.74.

2. Find the pH of 0.2 M CH₃COONa + 0.1 M CH₃COOH (pKa = 4.74).

pH = 4.74 + log(0.2/0.1) = 4.74 + 0.30 = 5.04.

3. Find the pH of a buffer of 0.1 M NH₄OH and 0.1 M NH₄Cl (pKb = 4.74).

pOH = pKb + log(0.1/0.1) = 4.74. pH = 14 − 4.74 = 9.26.

4. 100 mL of the buffer in example 1 (10 mmol acid, 10 mmol salt) gets 1 mmol HCl. Find the new pH.

H⁺ turns 1 mmol acetate into acid: salt = 9, acid = 11 mmol. pH = 4.74 + log(9/11) = 4.74 − 0.087 = 4.65. Only 0.09 down (water would drop from 7 to 2).

5. The solubility of AgCl is 1.34 × 10⁻⁵ mol/L. Find Ksp.

Ksp = s² = (1.34 × 10⁻⁵)² = 1.8 × 10⁻¹⁰.

6. Ksp of CaF₂ is 3.9 × 10⁻¹¹. Find its solubility.

Ksp = 4s³ → s³ = 9.75 × 10⁻¹² → s = 2.1 × 10⁻⁴ mol/L.

7. Will a precipitate form if equal volumes of 2 × 10⁻⁴ M AgNO₃ and 2 × 10⁻⁴ M NaCl are mixed? Ksp(AgCl) = 1.8 × 10⁻¹⁰.

After mixing each halves: [Ag⁺] = [Cl⁻] = 10⁻⁴ M. Q = 10⁻⁸. Q > Ksp, so AgCl precipitates.

8. Find the solubility of AgCl in 0.1 M NaCl (Ksp = 1.8 × 10⁻¹⁰).

[Cl⁻] ≈ 0.1 M from NaCl. s × 0.1 = 1.8 × 10⁻¹⁰ → s = 1.8 × 10⁻⁹ mol/L, about 7 400 times less than in water.

Common mistakes

Practice quiz

1. Which pair makes an acidic buffer?
2. When [salt] = [acid] in an acidic buffer, pH equals:
3. For CaF₂, Ksp in terms of s is:
4. A precipitate forms when:
5. Adding NaCl to a saturated AgCl solution:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What is a buffer solution?

A solution of a weak acid and its salt (or a weak base and its salt) that keeps its pH nearly constant when a little acid or base is added.

What is the relation between solubility and solubility product?

For AₓBᵧ, Ksp = xˣyʸ s^(x+y). For AB, Ksp = s²; for AB₂, Ksp = 4s³.

How does the common ion effect change solubility?

Adding an ion already in the equilibrium pushes it back, so less salt dissolves.

Where this is taught

CBSE (India)Class 11Equilibrium
USA (Common Core, NGSS, AP)Grade 11Equilibrium
USA (Common Core, NGSS, AP)Grade 11Acids and Bases
South Korea고등학교 2학년Acid-base equilibria
South Korea고등학교 3학년Reaction enthalpy and equilibrium
China高二Selective 1 Ch.3 Ionic equilibria

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