⚗️ Visual Chemistry
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Ionic Equilibrium & pH Calculations

25min Part 2 / Ch4 / Lesson 3
Prerequisites: 1-2-4 , 2-4-2

Objectives

  • Understand the ion-product of water Kw and use it in pH calculations
  • Calculate pH of weak acid/base solutions using Ka and Kb
  • Explain how buffer solutions work

Ion-Product of Water

Water undergoes very slight autoionization:

H2O⇌H++OH−\mathrm{H_2O \rightleftharpoons H^+ + OH^-}

Ion-product of water Kw=[H+][OH−]=1.0×10−14K_w = [\mathrm{H^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14} (at 25°C)

In pure water: [H+]=[OH−]=1.0×10−7 mol/L[\mathrm{H^+}] = [\mathrm{OH^-}] = 1.0 \times 10^{-7}\,\mathrm{mol/L} → pH = 7

KwK_w is constant at a given temperature. Adding acid increases [H+][\mathrm{H^+}] and decreases [OH−][\mathrm{OH^-}], but the product always equals 10−1410^{-14}.

Weak Acid Equilibrium

Dissociation of a weak acid HA:

HA⇌H++A−\mathrm{HA \rightleftharpoons H^+ + A^-}

Acid dissociation constant Ka=[H+][A−][HA]K_a = \dfrac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}

Larger KaK_a → stronger acid (more dissociation).

AcidFormulaKaK_apKa\mathrm{p}K_a
Acetic acidCH₃COOH1.8×10−51.8 \times 10^{-5}4.74
Carbonic acid (1st)H₂CO₃4.3×10−74.3 \times 10^{-7}6.37
PhenolC₆H₅OH1.0×10−101.0 \times 10^{-10}10.0

Representative Ka values (25°C)

pH of a Weak Acid

Find the pH of 0.10 mol/L acetic acid (Ka = 1.8×10⁻⁵)

CH3COOH⇌H++CH3COO−\mathrm{CH_3COOH \rightleftharpoons H^+ + CH_3COO^-}

Since the degree of dissociation α is small, [HA]≈c[\mathrm{HA}] \approx c (initial concentration).

Let [H+]=x[\mathrm{H^+}] = x:

Ka=x2c−x≈x2cK_a = \frac{x^2}{c - x} \approx \frac{x^2}{c}

x=Ka⋅c=1.8×10−5×0.10=1.8×10−6x = \sqrt{K_a \cdot c} = \sqrt{1.8 \times 10^{-5} \times 0.10} = \sqrt{1.8 \times 10^{-6}}

x=1.34×10−3 mol/Lx = 1.34 \times 10^{-3}\,\mathrm{mol/L}

pH=−log⁡(1.34×10−3)≈2.87\mathrm{pH} = -\log(1.34 \times 10^{-3}) \approx 2.87

Weak Base Equilibrium

For a weak base B:

B+H2O⇌BH++OH−\mathrm{B + H_2O \rightleftharpoons BH^+ + OH^-}

Kb=[BH+][OH−][B]K_b = \frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]}

Conjugate acid-base relationship: Ka⋅Kb=KwK_a \cdot K_b = K_w

Buffer Solutions

A buffer solution resists pH change when small amounts of acid or base are added.

A typical buffer is a mixture of weak acid HA and its salt NaA.

  • Adding acid (H⁺) → A⁻ reacts with H⁺ to form HA
  • Adding base (OH⁻) → HA reacts with OH⁻ to form A⁻ and H₂O

Henderson–Hasselbalch equation:

pH=pKa+log⁡[A−][HA]\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}

Buffer solutionHA ⇌ H⁺ + A⁻pH ≈ constant+H⁺A⁻ absorbs it+OH⁻HA reacts→ Equilibrium shifts to maintain pH
How a buffer works

Check Your Understanding

Q1 What is [H⁺] in pure water at 25°C?

Q2 A larger Ka means the acid...

Q3 A buffer solution is made by combining...

Exercises

Q1. Find the pH of 0.050 mol/L acetic acid (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}).

Solution

[H+]=Ka⋅c=1.8×10−5×0.050[\mathrm{H^+}] = \sqrt{K_a \cdot c} = \sqrt{1.8 \times 10^{-5} \times 0.050}

=9.0×10−7=9.5×10−4 mol/L= \sqrt{9.0 \times 10^{-7}} = 9.5 \times 10^{-4}\,\mathrm{mol/L}

pH=−log⁡(9.5×10−4)≈3.02\mathrm{pH} = -\log(9.5 \times 10^{-4}) \approx 3.02

Q2. Find the pH of a buffer made by mixing equal volumes of 0.10 mol/L acetic acid and 0.10 mol/L sodium acetate.

Solution

Henderson–Hasselbalch equation:

pH=pKa+log⁡[CH3COO−][CH3COOH]\mathrm{pH} = \mathrm{p}K_a + \log\dfrac{[\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}

Equal volumes → [A−]=[HA][\mathrm{A^-}] = [\mathrm{HA}] → log⁡1=0\log 1 = 0

pH=pKa=−log⁡(1.8×10−5)=4.74\mathrm{pH} = \mathrm{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74