Four kinds of acid–base reaction
In net ionic equations, strong acids and bases are written as ions and weak acids and bases as molecules (U4.2). The equilibrium constant of each kind of reaction comes from Ka, Kb and Kw:
- Strong acid + strong base: H₃O⁺ + OH⁻ → 2H₂O, K = 1 ÷ Kw = 1.0 × 10¹⁴.
- Weak acid + strong base: HA + OH⁻ → A⁻ + H₂O, K = Ka ÷ Kw.
- Strong acid + weak base: H₃O⁺ + B → BH⁺ + H₂O, K = Kb ÷ Kw.
- Weak acid + weak base: HA + B ⇌ A⁻ + BH⁺, K = Ka·Kb ÷ Kw.
The first three have very large K. For CH₃COOH + OH⁻, for example, K = 1.8 × 10⁻⁵ ÷ 1.0 × 10⁻¹⁴ = 1.8 × 10⁹, so the reaction can be treated as complete. So for a mixture, do the stoichiometry of the reaction first, then consider the equilibrium of whatever remains.
What’s left after the reaction?
List the amounts before, the change, and after, and see what kind of mixture remains:
- Excess strong acid or strong base: the excess H₃O⁺ or OH⁻ sets the pH (U8.2).
- Only a weak acid or a weak base: use Ka or Kb to find the pH (U8.3).
- A weak acid and its conjugate base (or a weak base and its conjugate acid): this is a buffer, with [H₃O⁺] = Ka × [HA] ÷ [A⁻].
- Exact reaction: if only a conjugate base remains, the solution is basic; if only a conjugate acid remains, it is acidic.
Choose an acid and a base, change their amounts, and see what remains after the reaction and what kind of solution it is:
Buffers
A solution containing comparable amounts of a weak acid and its conjugate base (or a weak base and its conjugate acid) is a buffer缓冲溶液. Its pH changes very little when a small amount of strong acid or strong base is added, because the added H₃O⁺ or OH⁻ is converted into a weak acid or weak base:
To make a buffer from a weak acid and a strong base, use fewer moles of the strong base than of the weak acid, so that both HA and A⁻ remain. U8.8–U8.10 cover more properties of buffers.
Acidic, basic and neutral salts
When a salt dissolves, ask whether its ions react with water:
- Anions of strong acids (Cl⁻, NO₃⁻, ClO₄⁻) and cations such as those of group 1 and Ca²⁺ don’t react with water. So NaCl and KNO₃ solutions are neutral.
- The conjugate bases of weak acids (F⁻, CH₃COO⁻, ClO⁻) react with water to form OH⁻. So NaF solution is basic.
- The conjugate acid of a weak base (NH₄⁺) reacts with water to form H₃O⁺. So NH₄Cl solution is acidic.
Example 1: partly neutralizing HF with NaOH
30.0 mL of 0.200 M HF (Ka = 6.4 × 10⁻⁴) is mixed with 10.0 mL of 0.200 M NaOH. Write the net ionic equation, classify the solution after the reaction, and estimate its pH.
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Write the net ionic equation. HF is a weak acid, so it is written as a molecule; NaOH is a strong base, so it is written as OH⁻: HF(aq) + OH⁻(aq) → F⁻(aq) + H₂O(l). This is the ionization of HF (Ka) plus H⁺ + OH⁻ → H₂O (1/Kw), so:
K is huge, so the reaction can be treated as complete.
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Amounts (in mmol). With volumes in mL, mL × mol/L = mmol, which keeps the numbers simple:
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BCA table (in mmol). OH⁻ is the smaller amount, so it is limiting:
HF OH⁻ F⁻ Before 6.00 2.00 0 Change −2.00 −2.00 +2.00 After 4.00 0 2.00 -
Classify the solution. After the reaction, significant amounts of both the weak acid HF and its conjugate base F⁻ are present, so this is a buffer.
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Estimate the pH. Rearranging the Ka expression gives [H₃O⁺] = Ka × [HF] ÷ [F⁻]. Both are in the same volume, so the concentration ratio equals the mole ratio:
The Henderson–Hasselbalch equation (U8.9) gives the same: pH = pKa + log(2.00 ÷ 4.00) = 3.19 − 0.30 = 2.89.
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Note. Ka for HF is fairly large: [H₃O⁺] ≈ 1.3 × 10⁻³ M is not tiny next to [F⁻] = 0.050 M, so HF’s own ionization can’t be ignored completely. An exact calculation using charge balance gives pH 2.91, only 0.02 different.
Example 2: exactly neutralizing NH₃ with HCl
60.0 mL of 0.15 M NH₃ is mixed with 30.0 mL of 0.30 M HCl. Find the pH of the mixture (Kb for NH₃ = 1.8 × 10⁻⁵).
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Amounts (in mmol).
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The reaction: exactly complete. H₃O⁺ + NH₃ → NH₄⁺ + H₂O (in mmol):
H₃O⁺ NH₃ NH₄⁺ Before 9.0 9.0 0 Change −9.0 −9.0 +9.0 After 0 0 9.0 Only NH₄⁺ (and Cl⁻) remains. Cl⁻ is the conjugate base of a strong acid and doesn’t affect the pH; NH₄⁺ is a weak acid and sets the pH.
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Concentration of NH₄⁺. The total volume is 60.0 + 30.0 = 90.0 mL:
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Ka of NH₄⁺. For a conjugate acid–base pair, Ka × Kb = Kw:
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Set up an ICE table. NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ (concentrations in M):
NH₄⁺ NH₃ H₃O⁺ I 0.10 0 0 C −x +x +x E 0.10 − x x x -
Solve with the approximation. Ka is tiny, so x ≪ 0.10:
7.5 × 10⁻⁶ is far smaller than 0.10, so the approximation holds.
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Find the pH.
The solution is acidic after exact neutralization, because the NH₄⁺ formed is a weak acid.
- 2 mmol CH₃COOH + 1 mmol NaOH
1 mmol CH₃COOH and 1 mmol CH₃COO⁻ remain.
- 1 mmol CH₃COOH + 1 mmol NaOH
Exact reaction; only CH₃COO⁻ remains.
- 1 mmol CH₃COOH + 2 mmol NaOH
NaOH is in excess.
- 2 mmol NH₃ + 1 mmol HCl
1 mmol each of NH₃ and NH₄⁺ remains.
- 1 mmol HCl + 1 mmol NaCl
Cl⁻ is the conjugate base of a strong acid and can't use up H₃O⁺.
- 1 mmol HF + 1 mmol KF
A weak acid and its conjugate base.
- 2 mmol NaF + 1 mmol HCl
1 mmol HF forms and 1 mmol F⁻ remains.
- 1 mmol HNO₃ + 1 mmol NaNO₃
HNO₃ is a strong acid.
Common mistakes
Mistake 1: “When an acid and a base exactly neutralize each other, the solution is always neutral.” Only a strong acid and a strong base give a neutral solution. A weak acid exactly neutralized by a strong base leaves the conjugate base, so the solution is basic; a strong acid with a weak base gives an acidic solution.
Mistake 2: “A buffer has pH 7.” The pH of a buffer depends on the pKa of the weak acid and the ratio of the two components; it can be acidic or basic.
Mistake 3: “After mixing, use Ka with the original concentrations to find the pH.” Do the neutralization first, then consider the equilibrium of what remains (in the total volume).
Quick check
What is the net ionic equation for the reaction of CH₃COOH solution with KOH solution?
CH₃COOH is a weak acid, so it is written as a molecule; K⁺ is a spectator ion. C is the molecular equation, not the net ionic equation.
Which pair, mixed in equal amounts, gives a buffer?
NH₃ is a weak base and NH₄⁺ is its conjugate acid. The acids in A and B are strong.
3.0 mmol NH₃ is mixed with 1.0 mmol HCl. After the reaction, the solution:
2.0 mmol NH₃ and 1.0 mmol NH₄⁺ remain, so it is a buffer; there is more NH₃ than NH₄⁺, so the pH is above the pKa of NH₄⁺ (9.26).
Which salt gives an acidic solution in water?
NH₄⁺ is the conjugate acid of the weak base NH₃; NO₃⁻ doesn't react with water. B and D are basic, and A is neutral.
Equal amounts of HCOOH and NaOH are mixed. At 25 °C the resulting solution:
After the exact reaction only HCOO⁻ remains; it is the conjugate base of a weak acid and reacts with water to form OH⁻.
What is K for CH₃COOH + NH₃ ⇌ CH₃COO⁻ + NH₄⁺? (Ka = 1.8 × 10⁻⁵, Kb(NH₃) = 1.8 × 10⁻⁵)
K = Ka·Kb ÷ Kw = (1.8 × 10⁻⁵)² ÷ 1.0 × 10⁻¹⁴ = 3.2 × 10⁴. K is large, so the reaction is essentially complete.
AP-style free response
A student mixes 50.0 mL of 0.200 M formic acid, HCOOH (Ka = 1.8 × 10⁻⁴), with 25.0 mL of 0.100 M NaOH.
(a) Write the net ionic equation for the reaction.
(b) Calculate the amounts of HCOOH and HCOO⁻ after the reaction.
(c) Estimate the pH of the mixture.
(d) Is this solution a buffer? Write a net ionic equation showing how it uses up a small amount of added HCl.
(e) More 0.100 M NaOH is added until the HCOOH has just completely reacted. What additional volume is needed? Calculate the pH at that point.
Model answer and scoring
(a) HCOOH(aq) + OH⁻(aq) → HCOO⁻(aq) + H₂O(l).
(b) n(HCOOH) = 10.0 mmol and n(OH⁻) = 2.50 mmol. After the reaction there are 7.50 mmol HCOOH and 2.50 mmol HCOO⁻.
(c) [H₃O⁺] ≈ Ka × 7.50 ÷ 2.50 = 5.4 × 10⁻⁴ M, so pH ≈ 3.27.
(d) Yes. The solution contains comparable amounts of the weak acid HCOOH and its conjugate base HCOO⁻. With HCl: HCOO⁻ + H₃O⁺ → HCOOH + H₂O.
(e) The remaining 7.50 mmol HCOOH needs 75.0 mL of NaOH. Then there are 10.0 mmol HCOO⁻ in a total of 150.0 mL, so [HCOO⁻] = 0.0667 M. Kb = Kw ÷ Ka = 5.56 × 10⁻¹¹ (keeping an extra digit in the intermediate result), so [OH⁻] ≈ √(5.56 × 10⁻¹¹ × 0.0667) = 1.92 × 10⁻⁶ M, pOH = 5.72 and pH = 8.28.
Scoring: (a) weak acid written as a molecule; (b) amounts after reaction; (c) pH from Ka and the ratio of the two components; (d) classification and equation; (e) volume, concentration of the conjugate base, and pH from Kb.
Summary
- Neutralizations involving a strong acid or strong base go to completion; do the stoichiometry first, then look at what remains.
- When a weak acid and its conjugate base (or a weak base and its conjugate acid) remain, the solution is a buffer: [H₃O⁺] = Ka × [HA] ÷ [A⁻].
- Salts of a strong acid and strong base are neutral, sodium salts of weak acids are basic, and ammonium salts are acidic.
AP tip: In a mixing problem, write the net ionic equation first, then list the amounts before, the change, and after. To explain why a solution is a buffer, say that it contains “comparable amounts of a weak acid and its conjugate base,” and write the equation for its reaction with added H₃O⁺ or OH⁻.