pH of weak bases — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Base dissociation constant ($K_b$), the $K_a$-$K_b$-$K_w$ relationship, approximation method for hydroxide concentration, pH calculation for pure weak bases, percent ionization of weak bases, and pH of basic salt solutions.

You should already know: Autoionization of water and the $K_w=[H^{+}][O{H}^{-}]$ relationship. pH and pOH definition and conversion. Equilibrium ICE table setup.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is pH of weak bases?

Weak bases are alkaline compounds that only partially ionize in aqueous solution, unlike strong bases which dissociate 100% into ions. Calculating the pH of a weak base solution relies on equilibrium principles, since the low degree of ionization means we cannot directly get $[O{H}^{-}]$ from the initial base concentration the way we do for strong bases. Per the AP Chemistry Course and Exam Description (CED), this topic contributes approximately 7-10% of the total exam weight for Unit 8 (Acids and Bases), and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. Common exam contexts include calculating pH of a given weak base solution, comparing pH of weak vs. strong bases of equal concentration, calculating pH of basic salts from strong base-weak acid reactions, and relating $K_b$ to base strength. Unlike weak acid pH calculation, which focuses on $[H^{+}]$ directly, weak base problems almost always first solve for $[O{H}^{-}]$ via equilibrium, then convert to pH using water's autoionization.

2. The Base Dissociation Constant ($K_b$) and $K_a$-$K_b$-$K_w$ Relationship

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 solvent, so it is excluded from the equilibrium expression (activity = 1 for pure liquids). The base dissociation constant $K_b$ is defined as: $$K_b=\frac{[{\text{BH}}^{+}][{\text{OH}}^{-}]}{[\text{B}]}$$ $K_b$ is a direct measure of base strength: a larger $K_b$ means a stronger base (more ionization, higher $[O{H}^{-}]$ for the same initial concentration). Most weak bases have $K_b<1$, consistent with their partial ionization behavior.

A key relationship connects $K_b$ of a base to $K_a$ of its conjugate acid: for any conjugate acid-base pair, multiplying $K_a$ (of the acid) and $K_b$ (of the conjugate base) gives $K_w$, the autoionization constant of water at 25°C: $$K_a\times K_b=K_w=1.0\times 10^{-14}$$ Taking negative base-10 logs of both sides gives the simplified p-version: $$\text{p}{K}_a+\text{p}{K}_b=14(\text{at }25^{\circ }C)$$ This relationship is critical because AP problems often give $K_a$ of the conjugate acid and ask for $K_b$ to calculate pH.

Worked Example

Problem: The $K_a$ of the ammonium ion ($N{H}_4^{+}$, conjugate acid of ammonia $N{H}_3$) is $5.6\times 10^{-10}$ at 25°C. Calculate $K_b$ for ammonia.

1. Recall the core relationship for conjugate pairs: $K_a\times K_b=K_w=1.0\times 10^{-14}$ at 25°C.
2. Rearrange to isolate $K_b$: $K_b=\frac{K_w}{K_a}$
3. Substitute the given values: $K_b=\frac{1.0\times 10^{-14}}{5.6\times 10^{-10}}=1.8\times 10^{-5}$
4. Check for reasonableness: Ammonia is a weak base, so $K_b<1$, which matches our result.

Exam tip: If a problem gives $\text{p}{K}_a$ instead of $K_a$, just subtract $\text{p}{K}_a$ from 14 to get $\text{p}{K}_b$ directly; you do not need to convert $\text{p}{K}_a$ back to $K_a$ first, which saves time on MCQ.

3. pH Calculation for a Pure Weak Base Solution

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 $[O{H}^{-}]$, then convert to pH. Let initial base concentration be $[B{]}_{initial}=c$. The ICE table gives equilibrium concentrations: $[B]=c-x$, $[B{H}^{+}]=x$, $[O{H}^{-}]=x$, where $x=[O{H}^{-}]$. Substitute into the $K_b$ expression: $$K_b=\frac{(x)(x)}{c-x}=\frac{x^2}{c-x}$$ Because $K_b$ is very small for weak bases, $x<<c$, so we can approximate $c-x\approx c$, simplifying the expression to: $$x=[O{H}^{-}]\approx \sqrt{K_b\times c}$$ 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 $[O{H}^{-}]$, calculate $\text{pOH}=-\log [O{H}^{-}]$, then $\text{pH}=14-\text{pOH}$ at 25°C.

Worked Example

Problem: Calculate the pH of a 0.15 M solution of methylamine ($C{H}_{3}N{H}_2$), where $K_b=4.4\times 10^{-4}$ at 25°C.

1. Write the equilibrium reaction: $C{H}_{3}N{H}_2(aq)+H_{2}O(l)\rightleftharpoons C{H}_{3}N{H}_3^{+}(aq)+O{H}^{-}(aq)$
2. Set up the ICE table: initial $[C{H}_{3}N{H}_2]=0.15$ M, all other starting concentrations = 0; change: $-x$, $+x$, $+x$; equilibrium: $0.15-x$, $x$, $x$.
3. Apply the approximation: $x=\sqrt{K_b\times c}=\sqrt{(4.4\times 10^{-4})(0.15)}=8.1\times 10^{-3}$ M.
4. Check the 5% rule: $\frac{8.1\times 10^{-3}}{0.15}\times 100\%=5.4\%$, which is just over 5%, so we solve the quadratic: $x^2+4.4\times 10^{-4}x-6.6\times 10^{-5}=0$, giving $x=7.9\times 10^{-3}$ M.
5. Calculate pH: $\text{pOH}=-\log (7.9\times 10^{-3})=2.10$, $\text{pH}=14-2.10=11.9$.

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.

4. pH of Basic Salts

Basic salts are ionic compounds that dissolve in water to produce a basic solution, formed from the neutralization of a strong base and a weak acid. For example, sodium acetate ($NaC{H}_{3}COO$) dissociates completely into $N{a}^{+}$ (a spectator ion from the strong base NaOH, which does not react with water) and $C{H}_{3}CO{O}^{-}$ (the conjugate base of the weak acid acetic acid). The acetate anion acts as a weak base in solution, so we calculate its pH exactly like any other weak base, using the $K_b$ of the anion, which we get from the $K_a$ of the parent weak acid via the $K_a{K}_b=K_w$ relationship. The spectator cation from the strong base does not affect pH and can be ignored entirely.

Worked Example

Problem: Calculate the pH of a 0.25 M solution of sodium hypochlorite (NaOCl). The $K_a$ of hypochlorous acid (HOCl) is $3.5\times 10^{-8}$ at 25°C.

1. Complete dissociation of the salt: $NaOCl(s)\to N{a}^{+}(aq)+OC{l}^{-}(aq)$, so $[OC{l}^{-}{]}_{initial}=0.25$ M, and $N{a}^{+}$ is a spectator.
2. Write the base equilibrium for $OC{l}^{-}$: $OC{l}^{-}(aq)+H_{2}O(l)\rightleftharpoons HOCl(aq)+O{H}^{-}(aq)$. Calculate $K_b$: $K_b=\frac{K_w}{K_a}=\frac{1.0\times 10^{-14}}{3.5\times 10^{-8}}=2.9\times 10^{-7}$.
3. Approximate $[O{H}^{-}]=\sqrt{K_b\times c}=\sqrt{(2.9\times 10^{-7})(0.25)}=2.7\times 10^{-4}$ M.
4. Check 5% rule: $\frac{2.7\times 10^{-4}}{0.25}\times 100\%=0.11\%<5\%$, so approximation is valid.
5. Calculate pH: $\text{pOH}=-\log (2.7\times 10^{-4})=3.57$, $\text{pH}=14-3.57=10.4$.

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 anion that is the conjugate base of a weak acid.

5. Percent Ionization of Weak Bases

Percent ionization is the percentage of the initial weak base that has ionized to produce $O{H}^{-}$ at equilibrium. It is calculated as: $$\text{Percent ionization}=\frac{[O{H}^{-}{]}_{equilibrium}}{[B{]}_{initial}}\times 100\%$$ Percent ionization correlates with both base strength and solution dilution. For a given weak base, percent ionization increases as the solution becomes more dilute (lower initial concentration). This follows Le Chatelier's principle: increasing the volume (diluting the solution) shifts the equilibrium toward the side with more moles of solute (1 mole of base produces 2 moles of ions), so more base ionizes. AP questions often ask you to compare percent ionization of different bases, or calculate $K_b$ from percent ionization and pH.

Worked Example

Problem: A 0.10 M solution of an unknown weak base has a pH of 10.5 at 25°C. Calculate the percent ionization of the base.

1. Calculate pOH from pH: $\text{pOH}=14-10.5=3.5$.
2. Calculate $[O{H}^{-}]=10^{-\text{pOH}}=10^{-3.5}=3.2\times 10^{-4}$ M.
3. Substitute into the percent ionization formula: $\text{Percent ionization}=\frac{3.2\times 10^{-4}}{0.10}\times 100\%=0.32\%$.
4. Check reasonableness: A percent ionization of 0.32% is well below 5%, which confirms the base is weak, matching the problem description.

Exam tip: If you are asked to calculate $K_b$ from percent ionization, rearrange the percent ionization formula to get $x=[O{H}^{-}]=\left(\frac{\text{percent}}{100}\right)\times c$, then plug $x$ and $c$ into $K_b=\frac{x^2}{c-x}$ to solve directly for $K_b$.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Using $[O{H}^{-}]=c$ (equal to initial base concentration) for weak bases, like you do for strong bases. Why: Students confuse the 100% dissociation rule for strong bases with partial dissociation for weak bases, and skip the equilibrium calculation. Correct move: Always confirm if the base is weak or strong first; if weak, always use $K_b$ and ICE to calculate $[O{H}^{-}]$.
• Wrong move: Solving directly for $[H^{+}]$ instead of $[O{H}^{-}]$ when setting up the equilibrium for weak bases. Why: Students memorize weak acid pH calculation and replicate it incorrectly, leading to wrong exponents and a final pH that is far too low. Correct move: For any weak base equilibrium, always set up the ICE table to solve for $[O{H}^{-}]$ first, then convert to pH via pOH.
• Wrong move: Forgetting that the anion of a weak acid acts as a weak base when calculating pH of basic salts, and assuming the salt is neutral. Why: Students forget only salts from strong acid-strong base neutralization are neutral; conjugate bases of weak acids hydrolyze to produce $O{H}^{-}$. Correct move: For any salt, split into cation and anion; if the anion is the conjugate base of a weak acid, treat it as a weak base for pH calculation.
• Wrong move: Misremembering the $K_a$-$K_b$ relationship, and using $K_b=\frac{K_a}{K_w}$ instead of $K_b=\frac{K_w}{K_a}$. Why: Students skip writing the full relationship and flip the fraction from memory. Correct move: Always write the full relationship $K_a{K}_b=K_w$ first before rearranging, every time.
• Wrong move: Including liquid water in the $K_b$ equilibrium expression. Why: Students include all reactants out of habit, forgetting solvent activity is 1. Correct move: Always omit pure liquid water from any $K_a$ or $K_b$ expression for aqueous equilibria.

7. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following 0.10 M solutions has the highest pH at 25°C? A) Hydrocyanic acid (HCN, $K_a=4.9\times 10^{-10}$) B) Sodium cyanide (NaCN) C) Ammonia (NH${}_3$, $K_b=1.8\times 10^{-5}$) D) Ammonium chloride (NH${}_4$Cl)

Worked Solution: First, eliminate options A and D: A is a weak acid (pH < 7) and D is a salt of a strong acid and weak base, which forms an acidic solution (pH < 7). Next, calculate $K_b$ for CN${}^{-}$ (the conjugate base of HCN in sodium cyanide): $K_b=\frac{K_w}{K_a}=\frac{1.0\times 10^{-14}}{4.9\times 10^{-10}}=2.0\times 10^{-5}$. Compare this to $K_b$ of ammonia (option C), which is $1.8\times 10^{-5}$. For equal concentrations, a higher $K_b$ gives higher $[O{H}^{-}]$, so sodium cyanide has a higher pH than ammonia. Correct answer: B

Question 2 (Free Response)

Trimethylamine ($(C{H}_3{)}_{3}N$) is a weak base found in decomposing organic matter. A solution of trimethylamine has an initial concentration of 0.20 M, and $K_b=6.3\times 10^{-5}$ at 25°C. (a) Write the balanced equilibrium equation for the ionization of trimethylamine in water, and write the $K_b$ expression. (b) Calculate the pH of the 0.20 M solution. Show whether your approximation is valid. (c) A dilute solution of trimethylamine has an initial concentration of 0.0020 M. Will the percent ionization of trimethylamine be higher, lower, or the same as in the 0.20 M solution? Justify your answer.

Worked Solution: (a) Balanced equilibrium: $$(C{H}_3{)}_{3}N(aq)+H_{2}O(l)\rightleftharpoons (C{H}_3{)}_{3}N{H}^{+}(aq)+O{H}^{-}(aq)$$ $K_b$ expression: $$K_b=\frac{[(C{H}_3{)}_{3}N{H}^{+}][O{H}^{-}]}{[(C{H}_3{)}_{3}N]}$$

(b) Set up ICE table: equilibrium concentrations are $0.20-x$, $x$, $x$ for trimethylamine, protonated trimethylamine, and hydroxide respectively. Approximate $0.20-x\approx 0.20$, so $x=\sqrt{K_{b}c}=\sqrt{(6.3\times 10^{-5})(0.20)}=3.55\times 10^{-3}$ M. Check 5% rule: $\frac{3.55\times 10^{-3}}{0.20}\times 100\%=1.8\%<5\%$, so approximation is valid. Calculate pH: $\text{pOH}=-\log (3.55\times 10^{-3})=2.45$, $\text{pH}=14-2.45=11.5$.

(c) Percent ionization will be higher in the 0.0020 M dilute solution. By Le Chatelier's principle, diluting an equilibrium with more moles of solute on the product side (1 mole of base produces 2 moles of ions) shifts the equilibrium right, increasing the fraction of ionized base.

Question 3 (Application / Real-World Style)

Swimming pool pH must be maintained between 7.2 and 7.8 for safe use. Sodium hypochlorite (bleach) is added to pools to kill bacteria, which dissociates to give hypochlorite ions ($OC{l}^{-}$), a weak base. A pool technician adds enough NaOCl to give a total $OC{l}^{-}$ concentration of 0.0040 M. The $K_a$ of HOCl is $3.5\times 10^{-8}$ at 25°C. Calculate the pH of the pool water from hypochlorite addition alone, and state whether the pH falls within the safe range.

Worked Solution: First calculate $K_b$ for $OC{l}^{-}$: $K_b=\frac{K_w}{K_a}=\frac{1.0\times 10^{-14}}{3.5\times 10^{-8}}=2.86\times 10^{-7}$. Approximate $[O{H}^{-}]=\sqrt{K_{b}c}=\sqrt{(2.86\times 10^{-7})(0.0040)}=3.4\times 10^{-5}$ M. Check the 5% rule: $\frac{3.4\times 10^{-5}}{0.0040}\times 100\%=0.85\%<5\%$, so approximation is valid. Calculate pH: $\text{pOH}=-\log (3.4\times 10^{-5})=4.47$, $\text{pH}=14-4.47=9.5$. The calculated pH of ~9.5 is well above the safe range of 7.2-7.8, so the pool would require acid addition to lower pH to a safe level.

8. Quick Reference Cheatsheet

9. What's Next

This chapter gives you the core equilibrium skills needed to calculate the pH of any base-containing solution, which is a critical prerequisite for the next topics in Unit 8 (Acids and Bases): acid-base titrations and buffer solutions. For example, when calculating the pH at the equivalence point of a strong acid-weak base titration, you rely on the $K_a$-$K_b$ relationship from this chapter to find the pH of the conjugate acid product. For buffer solutions made from a weak base and its conjugate salt, you will use $K_b$ directly to calculate buffer pH, so mastering weak base pH calculation is non-negotiable for these topics. Beyond Unit 8, this topic supports solubility equilibria in Unit 7, where the pH of the solution changes the solubility of ionic compounds with basic anions.

pH of weak acids Acid-base titrations Buffer solutions Solubility equilibria