pH & pKa
pH
- pH is defined as -log[H3O+]
- Note that [H+] is acceptable to use for AP chemistry
- To calculate [H3O+] from pH we can use log–pH
pKa
- pKa is a measure of the acidity or basicity of a substance in solution
- pKa = -log(Ka)
- To calculate Ka from pKa we can use log–pH
The relationship between pH and pKa
- The pH of a solution can be related to the pKa of an acid using the Henderson-Hasselbalch equation
pH = pKa + log
- [A–] is the concentration of the conjugate base
- [HA] is the concentration of the acid
pH curves, pH and pKa
- During a titration buffer regions occur when
- Strong acid - weak base
- Strong base - weak acid
- At this point, the buffer formed will resist changes in pH so the pH rises gradually as shown in the buffer region
- The half equivalence point is the stage of the titration at which exactly half the amount of weak acid has been neutralised
- At the half equivalence point, [acid] = [conjugate base] or [base] = [conjugate acid]
- For example during the titration of NaOH and ethanoic acid:
- CH3COOH + OH– CH3COO– + H2O
- CH3COOH is the acid
- CH3COO– is the conjugate base
The difference between pH and pKa
- pH is a measure of the acidity or basicity of an aqueous solution
- A pH below 7.00 indicates a substance is acidic
- A pH above 7.00 indicates a substance is basic
- pKa is used to show the strength of an acid
- A low pKa value indicates a strong acid
- A high pKa value indicates a weak acid
Worked example
Calculate the pKa of a 0.020 M solution of a weak acid with a pH value of 4.80.
Answer:
Step 1: Write the expression for Ka
- Ka =
Step 2: Calculate [H3O+] and make an ice chart to calculate [HA]
- [H3O+] = 10-pH = 10-4.80 = 1.58 x 10-5
HA | + | H2O (l) | A– (aq) | + | H3O+ | ||
Initial Conc. | 0.020 | - | 0.0 | 0.0 | |||
Change | 0.020 - 1.58 x 10-5 | - | + 1.58 x 10-5 | + 1.58 x 10-5 | |||
Equilibrium | 0.0199842 | - | 1.58 x 10-5 | 1.58 x 10-5 |
Step 3: Calculate Ka
- Ka = = 1.25 x 10-8
Step 4: Calculate pKa
- pKa = -log(1.25 x 10–8)
- pKa = 7.90