Weak Acid and Base Equilibria (Ka, Kb)弱酸弱碱的平衡（Ka、Kb）

Big Idea · Structure and Properties (SAP)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 8~50 min

Learning goals

  • Write Ka expressions for weak acids and Kb expressions for weak bases, and use ICE tables to find the pH of their solutions
  • Decide whether the approximation x ≈ √(K·c) is valid, calculate percent ionization, and explain why it increases on dilution
  • Use Ka·Kb = Kw to find Kb of a conjugate base from Ka of its acid, and compare the strengths of conjugate acids and bases

Before you start

  • Strong and weak acids and bases (2.12.2)
  • Calculating equilibrium concentrations (ICE tables) (U7.7)
  • pH and pOH of strong acids and bases (U8.2)

Ionization of a weak acid and Ka

Only a small fraction of a weak acid reacts with water, so most of it stays as un-ionized molecules. The equilibrium constant for this reaction is the acid ionization constant酸电离常数 Ka:

HA+HX2O⇌HX3OX++AX−Ka=[HX3OX+][AX−][HA]\begin{gathered} \ce{HA + H2O <=> H3O+ + A-} \\ K_a = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]} \end{gathered}

Water is a pure liquid, so it is left out of the expression. The larger Ka is, the stronger the acid. pKa = −log Ka, so the smaller pKa is, the stronger the acid.

A weak base reacts with water to form OH⁻, and its equilibrium constant is the base ionization constant碱电离常数 Kb:

B+HX2O⇌BHX++OHX−Kb=[BHX+][OHX−][B]\begin{gathered} \ce{B + H2O <=> BH+ + OH-} \\ K_b = \frac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]} \end{gathered}

Finding the pH with an ICE table

For a weak acid with initial concentration c, let x be the amount that ionizes. At equilibrium [H₃O⁺] = [A⁻] = x and [HA] = c − x (ignoring water’s ionization):

Ka=x2c−x≈x2cx≈Ka⋅c\begin{gathered} K_a = \frac{x^2}{c - x} \approx \frac{x^2}{c} \\ x \approx \sqrt{K_a \cdot c} \end{gathered}

The approximation requires x to be much smaller than c. The usual test is x ÷ c < 5%; otherwise, solve the quadratic (U7.7).

Percent ionization电离度 = x ÷ c × 100%. On dilution c gets smaller, so the percent ionization, about √(Ka ÷ c), gets larger, even though [H₃O⁺] still decreases.

Choose a weak acid or base and drag the concentration. Compare the orders of magnitude of the species, and watch how the percent ionization changes with concentration:

Try it: what is in a weak acid or weak base solution?

Conjugate pairs: Ka·Kb = Kw

HA and A⁻ form a conjugate acid–base pair共轭酸碱. Adding the ionization of HA to the reaction of A⁻ with water gives exactly the autoionization of water, so (U7.6):

Ka(HA)⋅Kb(AX−)=KwpKa+pKb=14.00  (25 °C)\begin{gathered} K_a(\ce{HA}) \cdot K_b(\ce{A-}) = K_w \\ \text{p}K_a + \text{p}K_b = 14.00 \;(25\,°\text{C}) \end{gathered}

The stronger the acid, the weaker its conjugate base. The conjugate bases of strong acids (Cl⁻, NO₃⁻ and so on) are too weak to react with water.

Example 1: pH of 0.20 M HF

For HF, Ka = 6.4 × 10⁻⁴ (OpenStax). Find the pH and percent ionization of 0.20 M HF.

Step-by-step solution

  1. Set up an ICE table. HF is a weak acid, so only a small fraction ionizes. Let x be the HF that ionizes (concentrations in M; H⁺ from water is negligible):

    HFH⁺F⁻
    I0.2000
    C−x+x+x
    E0.20 − xxx

  2. Write Ka and substitute.

    Ka=[HX+][FX−][HF]=x20.20−x=6.4×10−4\begin{aligned} K_a &= \frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]} \\ &= \frac{x^2}{0.20 - x} = 6.4 \times 10^{-4} \end{aligned}

  3. Try the approximation first. Assume x ≪ 0.20, so the denominator is 0.20:

    x≈6.4×10−4×0.20=1.28×10−4=1.13×10−2\begin{aligned} x &\approx \sqrt{6.4 \times 10^{-4} \times 0.20} \\ &= \sqrt{1.28 \times 10^{-4}} = 1.13 \times 10^{-2} \end{aligned}

  4. Test the approximation (the 5% rule).

    1.13×10−20.20×100%=5.7%>5%\frac{1.13 \times 10^{-2}}{0.20} \times 100\% = 5.7\% > 5\%

    That is over 5%, so the approximation isn’t good enough and the quadratic must be solved.

  5. Rearrange into a quadratic. Multiply both sides by (0.20 − x) and collect terms (keep the symbol Ka for now and substitute its value at the end):

    x2=Ka (0.20−x)x2+Ka x−0.20Ka=0\begin{gathered} x^2 = K_a\,(0.20 - x) \\ x^2 + K_a\,x - 0.20K_a = 0 \end{gathered}

  6. Use the quadratic formula. a = 1, b = Ka = 6.4 × 10⁻⁴ and c = −0.20Ka = −1.28 × 10⁻⁴:

    b2=4.1×10−7−4ac=5.12×10−4b2−4ac=5.124×10−4\begin{aligned} b^2 &= 4.1 \times 10^{-7} \\ -4ac &= 5.12 \times 10^{-4} \\ b^2 - 4ac &= 5.124 \times 10^{-4} \end{aligned} x=−6.4×10−4+0.022642=1.10×10−2 M\begin{aligned} x &= \frac{-6.4 \times 10^{-4} + 0.02264}{2} \\ &= 1.10 \times 10^{-2}\ \text{M} \end{aligned}

  7. Find the pH and percent ionization. x is [H⁺]:

    pH=−log⁡(1.10×10−2)=1.96\text{pH} = -\log(1.10 \times 10^{-2}) = \boxed{1.96}

    Percent ionization is the ionized HF as a percentage of the starting HF:

    1.10×10−20.20×100%=5.5%\frac{1.10 \times 10^{-2}}{0.20} \times 100\% = \boxed{5.5\%}

  8. Compare with the approximation. The approximation gives pH = −log(1.13 × 10⁻²) = 1.95, only 0.01 off, but by the 5% rule the exact answer should be shown.

Example 2: Kb from pH

At 25 °C, 0.050 M methylamine, CH₃NH₂, has pH 11.65. Find Kb for methylamine.

Step-by-step solution

  1. Get [OH⁻] from the pH (in M). Methylamine is a weak base, so start with the pOH:

    pOH=14.00−11.65=2.35[OHX−]=10−2.35=4.47×10−3\begin{aligned} \text{pOH} &= 14.00 - 11.65 = 2.35 \\ [\ce{OH-}] &= 10^{-2.35} = 4.47 \times 10^{-3} \end{aligned}

    Keep an extra digit in this intermediate result and round at the end.

  2. Set up an ICE table. CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH⁻. This time x is known: x = [OH⁻] = 4.47 × 10⁻³ (concentrations in M):

    CH₃NH₂CH₃NH₃⁺OH⁻
    I0.05000
    C−x+x+x
    E0.050 − xxx

  3. Find [CH₃NH₂] at equilibrium.

    0.050−0.00447=0.0455 M0.050 - 0.00447 = 0.0455\ \text{M}

  4. Write Kb and substitute.

    Kb=[CHX3NHX3X+][OHX−][CHX3NHX2]=(4.47×10−3)20.0455=2.00×10−50.0455=4.4×10−4\begin{aligned} K_b &= \frac{[\ce{CH3NH3+}][\ce{OH-}]}{[\ce{CH3NH2}]} \\ &= \frac{(4.47 \times 10^{-3})^2}{0.0455} \\ &= \frac{2.00 \times 10^{-5}}{0.0455} = \boxed{4.4 \times 10^{-4}} \end{aligned}

    This matches OpenStax Appendix I.

  5. Note. The percent ionization is 0.00447 ÷ 0.050 ≈ 9%, over 5%, so the denominator must be 0.050 − x = 0.0455, not 0.050 (which would give 4.0 × 10⁻⁴).

Practice: using x ≈ √(K·c), is x ÷ c below 5%?
  1. Ka = 1.8 × 10⁻⁵, c = 0.10 M
  2. Ka = 6.4 × 10⁻⁴, c = 0.010 M
  3. Ka = 4.9 × 10⁻¹⁰, c = 0.0010 M
  4. Ka = 1.2 × 10⁻², c = 0.10 M
  5. Ka = 2.9 × 10⁻⁸, c = 0.050 M
  6. Ka = 1.8 × 10⁻⁴, c = 0.50 M
  7. Kb = 4.4 × 10⁻⁴, c = 0.020 M
  8. Kb = 1.8 × 10⁻⁵, c = 0.20 M

A quick test: when c ÷ K > 400, x ÷ c is below 5%.

Common mistakes

Mistake 1: “In 0.10 M weak acid, [H₃O⁺] = 0.10 M.” A weak acid ionizes only slightly. In 0.10 M CH₃COOH, [H₃O⁺] is only about 1.3 × 10⁻³ M.

Mistake 2: “Diluting a weak acid lowers its percent ionization.” On dilution the equilibrium shifts toward ionization (Q < Ka), so the percent ionization increases; but because the total concentration is smaller, [H₃O⁺] still decreases.

Mistake 3: “An acid with a larger Ka also has a stronger conjugate base.” It’s the opposite: Ka·Kb = Kw, so the larger Ka is, the smaller Kb of the conjugate base.

Quick check

What is the Ka expression for HN₃ (hydrazoic acid)?

Which of these acids, at the same concentration, has the highest pH? (Ka: HF 6.4 × 10⁻⁴, HCOOH 1.8 × 10⁻⁴, HCN 4.9 × 10⁻¹⁰, HNO₂ 4.6 × 10⁻⁴)

0.10 M CH₃COOH is diluted to 0.010 M. Which change is correct?

For HF, Ka = 6.4 × 10⁻⁴. What is Kb for F⁻?

The pH of 0.25 M HCOOH (Ka = 1.8 × 10⁻⁴) is about:

Ka(HNO₂) = 4.6 × 10⁻⁴ and Ka(HClO) = 2.9 × 10⁻⁸. Which statement is correct?

AP-style free response

AP-style free response

Hypochlorous acid, HClO, is a weak acid with Ka = 2.9 × 10⁻⁸ (OpenStax, 25 °C).

(a) Write the equation for the reaction of HClO with water, and its Ka expression.

(b) Calculate the pH of 0.10 M HClO, and verify that your approximation is reasonable.

(c) Calculate the percent ionization of HClO in this solution.

(d) Calculate Kb for ClO⁻.

(e) Calculate the pH of 0.10 M NaClO, and explain why it is basic.

Model answer and scoring

(a) HClO(aq) + H₂O(l) ⇌ H₃O⁺(aq) + ClO⁻(aq); Ka = [H₃O⁺][ClO⁻] ÷ [HClO].

(b) x ≈ √(2.9 × 10⁻⁸ × 0.10) = 5.4 × 10⁻⁵ M, so pH = 4.27. x ÷ c = 0.054%, far below 5%, so the approximation is reasonable.

(c) 5.4 × 10⁻⁵ ÷ 0.10 × 100% = 0.054%.

(d) Kb = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 2.9 × 10⁻⁸ = 3.4 × 10⁻⁷.

(e) ClO⁻ + H₂O ⇌ HClO + OH⁻; [OH⁻] ≈ √(3.4 × 10⁻⁷ × 0.10) = 1.9 × 10⁻⁴ M, pOH = 3.73 and pH = 10.27. ClO⁻ is the conjugate base of the weak acid HClO, so it takes protons from water and forms OH⁻; Na⁺ doesn’t affect the pH.

Scoring: (a) equation and expression; (b) approximation, pH and check; (c) percent ionization; (d) Ka·Kb = Kw; (e) OH⁻ from Kb, pH, and the reason it is basic.

Summary

  • Weak acids and bases ionize only partly: Ka = [H₃O⁺][A⁻] ÷ [HA] and Kb = [BH⁺][OH⁻] ÷ [B].
  • Use an ICE table with x ≈ √(K·c); the approximation holds when x ÷ c < 5%, otherwise solve the quadratic.
  • On dilution, percent ionization rises and [H₃O⁺] falls. For a conjugate pair, Ka·Kb = Kw: the stronger the acid, the weaker its conjugate base.

AP tip: When you approximate, show the check: “x ÷ c = …% < 5%.” When finding Ka or Kb from pH, find x first, then decide whether c can replace c − x in the denominator. To compare conjugate bases, state directly that the smaller Ka is, the stronger the conjugate base.

Enrichment (polyprotic weak acids): Polyprotic acids such as H₂CO₃ and H₃PO₄ ionize in steps, with Ka₁ ≫ Ka₂. To find the pH, usually only the first step matters; the H₃O⁺ from the second step is small enough to ignore. For example, H₂CO₃ has Ka₁ = 4.3 × 10⁻⁷ and Ka₂ = 4.7 × 10⁻¹¹ (OpenStax).

Practice this chapter: Unit 8 · Acids and Bases · 12 questionsPractice this topic: U8.3 · Weak Acid and Base Equilibria (Ka, Kb) · 5 questions

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