# Lecture 6: pH, pOH, and Weak Acid/Base Equilibria
## General Chemistry II
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## Learning Objectives
By the end of this lecture, students will be able to:
1. Calculate pH, pOH, [H3O+], and [OH-] and convert between them
2. Set up and solve equilibrium problems for weak acid and weak base solutions
3. Calculate the percent ionization of a weak acid or base
4. Determine the pH of polyprotic acid solutions
5. Solve problems involving the hydrolysis of salts
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## Lecture Content
### I. The pH Scale
The pH scale provides a convenient way to express the hydronium ion concentration of a solution. It is defined as pH = -log[H3O+], and the analogous quantity for hydroxide is pOH = -log[OH-]. At 25 degrees C, pH and pOH are related by pH + pOH = 14.00. To convert back from pH to concentration, use [H3O+] = 10^(-pH) and [OH-] = 10^(-pOH).
The pH scale typically ranges from 0 to 14, although values outside this range are possible for very concentrated solutions. A pH less than 7 indicates an acidic solution, a pH of exactly 7 indicates neutrality at 25 degrees C, and a pH greater than 7 indicates a basic solution. Each unit change in pH corresponds to a tenfold change in [H3O+].
The pKa and pKb scales work analogously: pKa = -log(Ka) and pKb = -log(Kb), with pKa + pKb = pKw = 14.00 at 25 degrees C. A smaller pKa means a stronger acid, and a smaller pKb means a stronger base.
A vertical pH scale from 0 to 14 with common substances labeled at their approximate pH values. pH 0: battery acid (1 M HCl). pH 1: stomach acid. pH 2: lemon juice. pH 3: vinegar. pH 4: tomato juice. pH 5: coffee. pH 6: milk. pH 7: pure water (neutral, marked with a bold line). pH 8: seawater. pH 9: baking soda. pH 10: milk of magnesia. pH 11: ammonia solution. pH 12: soapy water. pH 13: bleach. pH 14: 1 M NaOH. The left side of the scale is colored red (acidic) fading to blue (basic) on the right. Corresponding [H3O+] values are listed alongside each pH value.
### II. pH of Strong Acid and Strong Base Solutions
Because strong acids dissociate completely, calculating the pH is straightforward. For a monoprotic strong acid, [H3O+] equals the initial acid concentration C_acid, and pH = -log(C_acid). For instance, 0.010 M HCl gives [H3O+] = 0.010 M and pH = 2.00.
Strong bases also dissociate completely. For Group 1A hydroxides, [OH-] equals C_base, while for Group 2A hydroxides, [OH-] = 2 * C_base because each formula unit releases two hydroxide ions. To find pH, first calculate pOH from the hydroxide concentration and then use pH = 14.00 - pOH. A 0.010 M NaOH solution, for example, has [OH-] = 0.010 M, pOH = 2.00, and pH = 12.00.
### III. pH of Weak Acid Solutions
Calculating the pH of a weak acid solution requires solving an equilibrium problem. For the ionization HA + H2O <=> H3O+ + A-, the Ka expression is Ka = [H3O+][A-] / [HA]. Setting up an ICE table with initial concentration C_a for HA and zero for both H3O+ and A-, the changes are -x, +x, and +x respectively. At equilibrium, Ka = x^2 / (C_a - x).
If the 5% approximation is valid (that is, x is much smaller than C_a), this simplifies to Ka approximately equals x^2 / C_a, giving x = [H3O+] = sqrt(Ka * C_a) and pH = -log(sqrt(Ka * C_a)) = 1/2(pKa - log C_a). Always verify the approximation by checking that x / C_a < 0.05. If the approximation fails, solve the full quadratic equation x^2 + Ka*x - Ka*C_a = 0 using the quadratic formula.
### IV. pH of Weak Base Solutions
The approach for weak bases mirrors that of weak acids, except that you solve for [OH-] instead of [H3O+]. For the equilibrium B + H2O <=> BH+ + OH-, the Kb expression is Kb = [BH+][OH-] / [B]. Using the 5% approximation, [OH-] = x = sqrt(Kb * C_b). Then pOH = -log[OH-] and pH = 14.00 - pOH.
For example, a 0.15 M NH3 solution with Kb = 1.8 x 10^-5 gives [OH-] = sqrt(1.8 x 10^-5 * 0.15) = 1.6 x 10^-3 M. This yields pOH = 2.79 and pH = 11.21.
### V. Percent Ionization
Percent ionization measures the fraction of a weak acid that actually ionizes: percent ionization = ([H3O+]_eq / [HA]_initial) x 100%. This quantity depends on both Ka and the initial concentration of the acid. A key relationship to remember is that percent ionization increases as the initial concentration decreases. At lower concentrations, a greater fraction of the acid must ionize to satisfy the Ka expression. This is sometimes called the dilution effect. Notably, doubling the concentration of a weak acid does not double [H3O+]; it only increases by a factor of approximately sqrt(2).
A graph showing percent ionization (y-axis, 0 to 100%) vs. initial acid concentration (x-axis, logarithmic scale from 10^-4 to 1 M) for three weak acids of different Ka values (Ka = 10^-2, Ka = 10^-4, Ka = 10^-6). Each curve shows that percent ionization increases as concentration decreases. The strongest acid (Ka = 10^-2) has the highest percent ionization at any given concentration. All curves approach 100% ionization at very low concentrations.
### VI. Polyprotic Acids
Polyprotic acids can donate more than one proton, with each successive dissociation characterized by its own Ka value. For example, H2SO3 has Ka1 = 1.5 x 10^-2 for its first dissociation (H2SO3 <=> H+ + HSO3-) and Ka2 = 6.3 x 10^-8 for its second (HSO3- <=> H+ + SO3^2-). A fundamental principle is that Ka1 >> Ka2 >> Ka3, because removing a proton from an increasingly negative species requires progressively more energy.
For pH calculations involving polyprotic acids, use only Ka1 to determine [H3O+] and pH, since the second dissociation contributes negligibly to the total hydronium ion concentration. A useful shortcut is that the concentration of the fully deprotonated species, [A^2-], approximately equals Ka2. The major exception is sulfuric acid, which is strong in its first dissociation; for H2SO4, the second dissociation (Ka2) must be considered using an equilibrium calculation. Common diprotic acids include H2SO4, H2SO3, H2CO3, H2S, and H2C2O4 (oxalic acid). Common triprotic acids include H3PO4 and H3C6H5O7 (citric acid).
### VII. Hydrolysis of Salt Solutions
Salts of weak acids and strong bases produce basic solutions because the anion hydrolyzes: A- + H2O <=> HA + OH-. The relevant equilibrium constant is Kb(A-) = Kw / Ka(HA), and you solve for [OH-] using an ICE table with this Kb value.
Salts of strong acids and weak bases produce acidic solutions because the cation hydrolyzes: BH+ + H2O <=> B + H3O+. Here, Ka(BH+) = Kw / Kb(B), and you solve for [H3O+] using an ICE table.
When both ions hydrolyze, as in salts of weak acids and weak bases, the solution's acidity depends on the relative magnitudes of Ka for the cation and Kb for the anion. If Ka > Kb, the solution is acidic. If Kb > Ka, the solution is basic. If Ka equals Kb, the solution is approximately neutral.
Small, highly charged metal cations such as Al3+, Fe3+, and Cr3+ represent a special case. These ions polarize the water molecules coordinated around them, making those water molecules significantly more acidic: [M(H2O)6]^3+ + H2O <=> [M(OH)(H2O)5]^2+ + H3O+.
A diagram showing salt hydrolysis for three example salts. Panel A: NaCH3COO (basic salt). Shows CH3COO- reacting with water to form CH3COOH + OH-. The equilibrium expression Kb = Kw/Ka is displayed. Panel B: NH4Cl (acidic salt). Shows NH4+ reacting with water to form NH3 + H3O+. The equilibrium expression Ka = Kw/Kb is displayed. Panel C: NH4CH3COO (ambiguous salt). Shows both ions hydrolyzing and indicates that the relative magnitudes of Ka(NH4+) = 5.6 x 10^-10 and Kb(CH3COO-) = 5.6 x 10^-10 determine pH. Since Ka = Kb, the solution is approximately neutral.
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