Acid & Base Calculations
pH
- The acidity of an aqueous solution depends on the number of H3O+ ions in solution
- pH is defined as:
pH = -log [H3O+]
- Where [H3O+] is the concentration of H3O+ ions in mol dm–3
- Similarly, the concentration of H+ of a solution can be calculated if the pH is known by rearranging the above equation to:
[H3O+] = 10-pH
- The pH scale is a logarithmic scale with base 10
- This means that each value is 10 times the value below it
- For example, pH 5 is 10 times more acidic than pH 6
- pH values are usually given to 2 decimal places
pOH
- The basicity of an aqueous solution depends on the number of hydroxide ions, OH-, in solution
- pOH is defined as:
pOH = -log [OH-]
- Where [OH-] is the concentration of hydroxide ions in mol dm–3
- Similarly, the concentration of OH- of a solution can be calculated if the pH is known by rearranging the above equation to:
[OH-] = 10-pOH
- If you are given the concentration of a basic solution and need to find the pH, this can be done by:
[H3O+] = Kw / [OH-]
- Alternatively, if you are given the [OH-] and calculate the pOH, the pH can be found by:
pH = 14 - pOH
Worked Example
pH and H3O+ calculations
- Find the pH when the hydrogen ion concentration is 1.60 x 10-4 mol dm-3
- Find the hydrogen ion concentration when the pH is 3.10
Answers
Answer 1:
The pH of the solution is:
- pH = -log [H3O+]
- pH = -log 1.6 x 10-4
- pH = 3.80
- pH = -log [H3O+]
Answer 2:
The hydrogen concentration can be calculated by rearranging the equation for pH
- pH = -log [H3O+]
- [H3O+] = 10-pH
- [H3O+] = 10-3.10
- [H3O+] = 7.94 x 10-4 mol dm-3
Worked Example
pH calculations of a strong alkali
- Calculate the pH of 0.15 mol dm-3 sodium hydroxide, NaOH
- Calculate the hydroxide concentration of a solution of sodium hydroxide when the pH is 10.50
Answers
Sodium hydroxide is a strong base which ionises as follows:
NaOH (aq) → Na+ (aq) + OH– (aq)
Answer 1:
The pH of the solution is:
- [H+] = Kw ÷ [OH–]
- [H+] = (1 x 10-14) ÷ 0.15 = 6.66 x 10-14
- pH = -log[H+]
- pH = -log 6.66 x 10-14 = 13.17
- [H+] = Kw ÷ [OH–]
Answer 2
Step 1: Calculate hydrogen concentration by rearranging the equation for pH
- pH = -log[H+]
- [H+] = 10-pH
- [H+] = 10-10.50
- [H+] = 3.16 x 10-11 mol dm-3
Step 2: Rearrange the ionic product of water to find the concentration of hydroxide ions
- Kw = [H+] [OH–]
- [OH–] = Kw ÷ [H+]
Step 3: Substitute the values into the expression to find the concentration of hydroxide ions
- Since Kw is 1 x 10-14 mol2 dm-6
- [OH–] = (1 x 10-14) ÷ (3.16 x 10-11)
- [OH–] = 3.16 x 10-4 mol dm-3
- Since Kw is 1 x 10-14 mol2 dm-6
Ka, pKa, Kb and pKb
- In reactions of weak acids and bases, we cannot make the same assumptions as for the ionisation of strong acids and bases
- For a weak acid and its conjugate base, we can use the equation:
Kw = Ka Kb
- By finding the -log of these, we can use:
pKw = pKa + pKb
- Remember, to convert these terms you need to use:
pKa = -logKa Ka= 10–pKa
pKb = -logKb Kb= 10–pKb
- The assumptions we must make when calculating values for Ka, pKa, Kb and pKb are:
- The initial concentration of acid ≈ the equilibrium concentration of acid
- [A-] = [H3O+]
- There is negligible ionisation of the water, so [H3O+] is not affected
- The temperature is 25 °C
Worked Example
Calculate the acid dissociation constant, Ka, at 298 K for a 0.20 mol dm-3 solution of propanoic acid with a pH of 4.88.
Answer
Step 1: Calculate [H3O+] using
- [H3O+] = 10-pH
- [H3O+] = 10-4.88
- [H3O+] = 1.3182 x 10-5
- [H3O+] = 10-pH
Step 2: Substitute values into Ka expression (include image)
- Ka = [H3O+]2 / [CH3CH2COOH]
- Ka = (1.3182 x 10-5) 2 / 0.2
- Ka = 8.70 × 10-10 mol dm-3
- Ka = [H3O+]2 / [CH3CH2COOH]
Worked Example
A 0.035 mol dm-3 sample of methylamine (CH3NH2) has pKb value of 3.35 at 298 K. Calculate the pH of methylamine.
Answer
Step 1: Calculate the value for Kb using
-
- Kb = 10–pKb
- Kb= 10-3.35
- Kb = 4.4668 x 10-4
- Kb = 10–pKb
Step 2: Substitute values into Kb expression to calculate [OH-]
-
- Kb = [OH-]2 / [CH3NH2]
- 4.4668 x 10-4 = [OH-]2 / 0.035
- [OH-] = √(4.4668 x 10-4 x 0.035)
- [OH-] = 3.9539 x 10-3
Step 3: Calculate the pH
-
- [H+] = Kw ÷ [OH–]
- [H+] = (1 x 10-14) ÷ 3.9539 x 10-3
- [H+] = 2.5290 x 10-12
- pH = -log [H+]
- pH = 2.5290 x 10-12
- pH = 11.60 to 2 decimal places
- [H+] = Kw ÷ [OH–]
OR
Step 3: Calculate pOH and therefore pH
-
- pOH = -log [OH-]
- pOH = -log 3.9539 x 10-3
- pOH = 2.4029
- pOH = -log [OH-]
-
- pH = 14 - pOH
- pH = 14 - 2.4029
- pH = 11.60 to 2 decimal places
- pH = 14 - pOH