pH of weak bases — AP Chemistry
1. Base Dissociation Constant ($K_b$) and $K_a$-$K_b$-$K_w$ Relationship ★★☆☆☆ ⏱ 3 min
When a weak base $\text{B}$ dissolves in water, it accepts a proton from water, following the equilibrium:
\text{B}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{BH}^+(aq) + \text{OH}^-(aq)
Water is the pure solvent, so it is excluded from the equilibrium expression (activity = 1 for pure liquids). The $K_b$ expression is:
K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}
For any conjugate acid-base pair at 25°C, the product of the acid dissociation constant of the acid and the base dissociation constant of the conjugate base equals $K_w$, the autoionization constant of water:
K_a \times K_b = K_w = 1.0 \times 10^{-14}
\text{p}K_a + \text{p}K_b = 14 \quad (\text{at } 25^\circ C)
Exam tip: If a problem gives $\text{p}K_a$ instead of $K_a$, subtract $\text{p}K_a$ from 14 to get $\text{p}K_b$ directly, saving time on multiple-choice questions.
2. pH Calculation for a Pure Weak Base Solution ★★★☆☆ ⏱ 4 min
To find the pH of a solution of a pure weak base with known initial concentration and $K_b$, we use an ICE (Initial, Change, Equilibrium) table to find equilibrium $[OH^-]$, then convert to pH. If the initial base concentration is $c$, the ICE table gives equilibrium concentrations: $[B] = c - x$, $[BH^+] = x$, $[OH^-] = x$, where $x = [OH^-]$. Substituting into the $K_b$ expression gives:
K_b = \frac{x^2}{c - x}
Because $K_b$ is very small for weak bases, $x << c$, so we can approximate $c - x \approx c$, simplifying the expression to:
x = [OH^-] \approx \sqrt{K_b \times c}
After calculating $x$, we check the 5% rule: if $\frac{x}{c} \times 100\% < 5\%$, the approximation is valid. If not, we solve the quadratic equation for the exact value of $x$. Once we have $[OH^-]$, calculate $\text{pOH} = -\log[OH^-]$, then $\text{pH} = 14 - \text{pOH}$ at 25°C.
Exam tip: AP exam graders accept answers within 0.1 pH unit of the correct value, even if you use the approximation when percent ionization is 5-6%, but always explicitly state whether your approximation is valid to earn full points on FRQ.
3. pH of Basic Salts ★★★☆☆ ⏱ 3 min
Basic salts are ionic compounds formed from the neutralization of a strong base and a weak acid. They dissolve completely in water to release a spectator cation (from the strong base, which does not react with water) and an anion (the conjugate base of the weak acid, which acts as a weak base in solution). We calculate pH for basic salts exactly the same way as for any other weak base.
Exam tip: Always identify spectator ions first when solving basic salt pH problems: all group 1 and heavy group 2 metal cations from strong bases do not affect pH, so you only need to focus on the conjugate base anion.
4. Percent Ionization of Weak Bases ★★☆☆☆ ⏱ 2 min
Percent ionization is the percentage of the initial weak base that has ionized to produce $OH^-$ at equilibrium. It is calculated as:
\text{Percent ionization} = \frac{[OH^-]_{equilibrium}}{[B]_{initial}} \times 100\%
Percent ionization correlates with both base strength and solution dilution. For a given weak base, percent ionization increases as the solution becomes more dilute. This follows Le Chatelier's principle: increasing the volume (diluting) shifts equilibrium toward the side with more moles of solute (1 mole of base produces 2 moles of ions), so more base ionizes.
Exam tip: If you are asked to calculate $K_b$ from percent ionization, rearrange the formula to get $x = [OH^-] = \left(\frac{\text{percent}}{100}\right) \times c$, then plug $x$ and $c$ into $K_b = \frac{x^2}{c-x}$ to solve directly for $K_b$.
Common Pitfalls
Why: Students confuse the 100% dissociation rule for strong bases with partial dissociation for weak bases, and skip the required equilibrium calculation
Why: Students memorize weak acid pH calculation and replicate it incorrectly, leading to wrong exponents and a final pH that is far too low
Why: Students forget only salts from strong acid-strong base neutralization are neutral; conjugate bases of weak acids hydrolyze to produce $OH^-$
Why: Students skip writing the full relationship and flip the fraction from memory
Why: Students include all reactants out of habit, forgetting pure solvent activity is 1