pH & pKa (College Board AP® Chemistry)
Study Guide
Written by: Philippa Platt
Reviewed by: Stewart Hird
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
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