Notes
pH basics
- $pH = -\log[H_3O^+]$, $pOH = -\log[OH^-]$, and at 25 °C, $pH + pOH = 14$ and $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$.
- pH below 7 is acidic; above 7 is basic. Each pH unit is a 10× change in $[H_3O^+]$.
Strong and weak
- Strong acids ionize completely: HCl, HBr, HI, $\mathrm{HNO_3}$, $\mathrm{H_2SO_4}$, $\mathrm{HClO_4}$. Strong bases: Group 1 hydroxides and $Ca$, $Sr$, $Ba$ hydroxides.
- Weak acids ionize partly: $\displaystyle K_a = \frac{[H_3O^+][A^-]}{[HA]}$. A larger $K_a$ (smaller $pK_a$) means a stronger acid.
- For a weak acid, $\displaystyle [H_3O^+] \approx \sqrt{K_a C}$ when ionization is small.
- Conjugate pairs: $K_a \times K_b = K_w$. The stronger the acid, the weaker its conjugate base.
- Acid strength from structure: more electronegative atoms near the O–H and more stable conjugate bases make stronger acids.
Titrations
- At the equivalence point, moles of acid equal moles of base. Strong acid + strong base: pH 7. Weak acid + strong base: pH above 7. Weak base + strong acid: pH below 7.
- At the half-equivalence point of a weak acid titration, $pH = pK_a$.
- Choose an indicator whose color change ($pK_a$ of the indicator) is near the equivalence pH.
Buffers
A buffer has a weak acid and its conjugate base (or a weak base and its conjugate acid) in similar amounts. It resists pH changes because the base neutralizes added acid and the acid neutralizes added base.
- Henderson–Hasselbalch: $\displaystyle pH = pK_a + \log\frac{[A^-]}{[HA]}$.
- Buffer capacity is greater with higher concentrations of both components.
- When $pH < pK_a$, the acid form HA dominates; when $pH > pK_a$, $\mathrm{A^-}$ dominates.
Topics in this unit
Each topic has its own short notes and practice.
- 8.1Introduction to Acids and Bases
- 8.2pH and pOH of Strong Acids and Bases
- 8.3Weak Acid and Base Equilibria
- 8.4Acid-Base Reactions and Buffers
- 8.5Acid-Base Titrations
- 8.6Molecular Structure of Acids and Bases
- 8.7pH and pKa
- 8.8Properties of Buffers
- 8.9Henderson-Hasselbalch Equation
- 8.10Buffer Capacity
- 8.11pH and Solubility
Practice
Strong acid: $[H_3O^+] = 0.010 = 10^{-2}$.
$pOH = 3$, so $pH = 14 - 3$.
HF only partly ionizes in water.
Remove one $\mathrm{H^+}$.
At half-equivalence, $[HA] = [A^-]$, so $pH = pK_a$.
Only the conjugate base remains, which makes the solution basic.
A weak acid with its conjugate base.
$5.0 + \log 10 = 6.0$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A 0.10 M solution of a weak acid HA has a pH of 3.00.
(a) Calculate $[H_3O^+]$ and $K_a$.
Show the answer and scoring guide
$[H_3O^+] = 1.0 \times 10^{-3}$ M. $\displaystyle K_a = \frac{(1.0 \times 10^{-3})^2}{0.10 - 0.0010} \approx 1.0 \times 10^{-5}$.
Check off the points you earned:
(b) Calculate the percent ionization.
Show the answer and scoring guide
$\displaystyle \frac{1.0 \times 10^{-3}}{0.10} \times 100 = 1.0\%$.
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(c) 25.0 mL of this acid is titrated with 0.10 M NaOH. What is the pH after 12.5 mL of NaOH is added? Explain.
Show the answer and scoring guide
That's the half-equivalence point (half the acid has been neutralized), so $pH = pK_a = -\log(1.0 \times 10^{-5}) = 5.00$.
Check off the points you earned: