12.1 Autoionization of Water, pH Scale & Strong vs Weak Acids/Bases

12.1 Autoionization of Water, pH Scale & Strong vs Weak Acids/Bases

Quick Summary: Water undergoes endothermic autoionization ($2\text{ H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-$) defined by $K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14}$ at 25 °C. Because autoionization absorbs heat, $K_w$ increases with temperature, lowering neutral water's pH below 7.00 at elevated temperatures. Strong acids and bases dissociate completely under water's leveling effect. Weak acids and bases establish partial dissociation equilibria ($K_a$, $K_b$), with percent ionization increasing upon dilution via Ostwald's dilution law. Conjugate acid-base pairs are linked by $K_a \times K_b = K_w$ and $\text{p}K_a + \text{p}K_b = 14.00$.

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1. Autoionization of Water & the Ion Product Constant ($K_w$)

Liquid water is amphiprotic, functioning as both proton donor and acceptor: $2\text{ H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq)$ Excluding the solvent ($a = 1$) yields the ion product constant of water ($K_w$): $K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ at } 25\ ^\circ\text{C}$ In pure water at 25 °C, stoichiometry dictates $[\text{H}_3\text{O}^+] = [\text{OH}^-] = \sqrt{K_w} = 1.0 \times 10^{-7}\text{ M}$, giving $\text{pH} = 7.00$.

Temperature Dependence of $K_w$

Autoionization is an endothermic process ($\Delta H^\circ > 0$). Elevating temperature shifts equilibrium to the right, increasing $K_w$:

• At $25\ ^\circ\text{C}$: $K_w = 1.0 \times 10^{-14} \implies [\text{H}_3\text{O}^+] = 1.0 \times 10^{-7}\text{ M} \implies \text{pH} = 7.00$
• At $37\ ^\circ\text{C}$ (body temperature): $K_w \approx 2.4 \times 10^{-14} \implies [\text{H}_3\text{O}^+] = \sqrt{2.4 \times 10^{-14}} \approx 1.55 \times 10^{-7}\text{ M} \implies \text{pH} \approx 6.81$

Crucial Exam Concept: Chemical neutrality is defined strictly by $[\text{H}_3\text{O}^+] = [\text{OH}^-]$, never by an invariant pH value of 7.00. Water at 37 °C with $\text{pH} = 6.81$ is neutral because hydronium and hydroxide concentrations remain identical.

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2. Quantitative pH/pOH Scales & Conversion Relationships

Sørensen defined the logarithmic acidity scales: $\text{pH} = -\log_{10}[\text{H}_3\text{O}^+] \quad \Longleftrightarrow \quad [\text{H}_3\text{O}^+] = 10^{-\text{pH}}$ $\text{pOH} = -\log_{10}[\text{OH}^-] \quad \Longleftrightarrow \quad [\text{OH}^-] = 10^{-\text{pOH}}$ Taking the negative logarithm of $K_w$ yields: $\text{pH} + \text{pOH} = \text{p}K_w = 14.00 \text{ at } 25\ ^\circ\text{C}$

The Acid-Base Conversion Circle

       [H3O+]  <===== Kw = [H3O+][OH-] =====>  [OH-]
         ^                                       ^
         |                                       |
 pH = -log[H3O+]                         pOH = -log[OH-]
 [H3O+] = 10^(-pH)                       [OH-] = 10^(-pOH)
         |                                       |
         v                                       v
        pH     <===== pH + pOH = 14.00 =====>   pOH

Significant Figures in Logarithms

The number of decimal places in the pH value (the mantissa) equals the number of significant figures in the hydronium concentration:

• If $[\text{H}_3\text{O}^+] = 3.5 \times 10^{-4}\text{ M}$ (2 sig figs), $\text{pH} = 3.46$ (2 decimal places).
• If $\text{pH} = 8.15$ (2 decimal places), $[\text{H}_3\text{O}^+] = 10^{-8.15} = 7.1 \times 10^{-9}\text{ M}$ (2 sig figs).

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3. Strong Acids and Strong Bases: Complete Ionization

Strong electrolytes dissociate completely ($\alpha = 100\%$) in water. Due to the leveling effect of water, $\text{H}_3\text{O}^+$ is the strongest acid and $\text{OH}^-$ is the strongest base that can exist in aqueous solution.

Strong Acids and Bases Master Reference

Most general chemistry courses (and the net ionic rules in Section 8.1) memorize six strong acids: HCl, HBr, HI, HNO₃, HClO₄, and H₂SO₄ (first proton). Chloric acid, HClO₃, is also usually treated as strong.

Direct pH Calculations

• For $0.025\text{ M }\text{HNO}_3$: $[\text{H}_3\text{O}^+] = 0.025\text{ M} \implies \text{pH} = -\log_{10}(0.025) = 1.60$.
• For $0.015\text{ M }\text{Ba(OH)}_2$: $[\text{OH}^-] = 2(0.015\text{ M}) = 0.030\text{ M} \implies \text{pOH} = 1.52 \implies \text{pH} = 12.48$.

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4. Weak Acids, Weak Bases & Ostwald's Dilution Law

Weak electrolytes establish dynamic partial ionization equilibria: $\text{HA}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{A}^-(aq)$ $K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$ $\text{B}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{BH}^+(aq) + \text{OH}^-(aq)$ $K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$

Percent Ionization & Dilution

$\% \text{ Ionization} = \frac{[\text{H}_3\text{O}^+]_{\text{eq}}}{[\text{HA}]_0} \times 100\%$ By Ostwald's dilution law, diluting a weak acid increases its percent ionization. Adding solvent increases volume $V$, making $Q < K_a$. The system shifts toward the side with more dissolved particles (products). Thus, a larger fraction of acid ionizes, even as total $[\text{H}_3\text{O}^+]$ decreases.

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5. Conjugate Pairs: $K_a \times K_b = K_w$

Summing acid dissociation ($\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$) and conjugate base hydrolysis ($\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-$) yields water's autoionization: $K_a \times K_b = K_w = 1.0 \times 10^{-14} \implies \text{p}K_a + \text{p}K_b = 14.00 \text{ at } 25\ ^\circ\text{C}$ The stronger an acid, the weaker its conjugate base. Conjugate bases of strong acids lack basicity in water and act as spectator ions.

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6. Worked Example: Weak Acid Equilibrium

Problem: Find $[\text{H}_3\text{O}^+]$, pH, and percent ionization of $0.150\text{ M}$ acetic acid ($\text{CH}_3\text{COOH}$, $K_a = 1.8 \times 10^{-5}$) at 25 °C.

Step 1: Set up equilibrium expression $\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{CH}_3\text{COO}^- \implies K_a = \frac{x^2}{0.150 - x} = 1.8 \times 10^{-5}$

Step 2: Solve with approximation Since $0.150 / (1.8 \times 10^{-5}) = 8333 \gg 400$, assume $0.150 - x \approx 0.150$: $x^2 = (0.150)(1.8 \times 10^{-5}) = 2.70 \times 10^{-6} \implies x = [\text{H}_3\text{O}^+] = 1.64 \times 10^{-3}\text{ M}$

Step 3: Calculate pH and percent ionization $\text{pH} = -\log_{10}(1.643 \times 10^{-3}) = 2.78$ $\% \text{ Ionization} = \frac{1.643 \times 10^{-3}\text{ M}}{0.150\text{ M}} \times 100\% = 1.10\%$ Valid because $1.10\% < 5\%$. (Carry the unrounded $x$ into the logarithm; rounding $x$ to $1.64 \times 10^{-3}$ first would give $2.79$.)