1. Acids and Bases (HL) - Lesson 8
Weak acid/base calculations

2. Refresh
Based on information in the table below, which acid is the strongest?
Acid pKa Ka
A. HA –
B. HB – 1 × 10–3
C. HC –
D. HD – 1 × 10–5

3. We Are Here

4. Lesson 7: Weak acid/base calculations
Objectives:
Perform calculations involving weak acids
Perform calculations involving weak bases

5. Recap WEAK ACIDS Ka = [H+][A-]/[HA] pKa = -log10(Ka)
Smaller Ka  weaker acid
Smaller pKa  stronger acid
WEAK BASES
Kb = [B+][OH-]/[BOH]
pKb = -log10(Kb)
Smaller Kb  weaker base
Smaller pKb  stronger base
Acid Ka
pKa
Hydronium ion, H3O+
1.00
0.00
Oxalic acid, HO2CCO2H
5.9x10-2
1.23
Hydrofluoric, HF
7.2x10-4
3.14
Methanoic, CHOOH
1.77x10-4
3.75
Ethanoic, CH3COOH
1.76x10-5
4.75
Phenol, C6H5OH
1.6x10-10
9.80
Base Kb
pKb
Diethylamine, (C2H5)2NH
1.3x10-3
2.89
Ethylamine, C2H5NH2
5.6x10-4
3.25
Methylamine, CH3NH2
4.4x10-4
3.36
Ammonia, NH3
1.8x10-5
4.74

6. We need to be able to solve problems such as:
What is the pH of an 1.50 M solution of weak acid, X?
What is the [OH-] of a solution of weak base, Y?
50 cm3 of a 0.1 M solution of acid X reacts with 25 cm3 of a 0.1 M solution of base Y, what is the resulting pH?
The pH of a M solution of weak acid Z is 5.4, what is it’s Ka and pKa?

7. We will need to use a variety of equations:
New(ish) today:
Ka × Kb = Kw = 1.00x10-14
pKa + pKb = pKw = 14.0
pH + pOH = pKw = 14.0
And from previous lessons:
pKa = -log10(Ka) pKb = -log10(Kb)
pH = -log10[H+] pOH = -log10[OH-]

8. Example 1: Calculation of [OH-]
What is the concentration of OH- ions in a mol dm-3 solution of ammonia (Kb = 1.8x10-5)? What % of the NH3 molecules have dissociated?
Since it is a weak base and the equilibrium is to the right we assume that at equilibrium, [NH3] is the same as stated in the question. So:
Kb = [NH4+][OH-]/[NH3] Sub all known values into equation
1.8x10-5 = [NH4+][OH-]/ Looks like there are 2 unknowns
However, since [NH4+] = [OH-]:
1.8x10-5 = [OH-]2 / Rearrange to make [OH-] the subject
[OH-] = √(1.8x10-5 x 0.500) Perform calculation
[OH-] = mol dm-3
% Dissociation = / x 100 = 0.60%

9. Example 2: Calculating pH
What is the pH of a mol dm-3 solution of oxalic acid (HOOCCOOH, Ka = 5.9x10-2), and how does the pH change on ten-fold dilution?
Again, assume [HA] is as stated in the question
Ka = [H+][A-] / [HA] Sub in all known values
5.9x10-2 = [H+][A-] / Looks like two unknowns, but isn’t really
This time, since oxalic acid produces two protons, [A-] = ½ [H+] so the expression becomes:
5.9x10-2 = ½ [H+]2 / Rearrange to make [H+] subject
[H+] = √((5.9x10-2 x 0.225) / 2) = mol dm-3
pH = -log10[H+] = -log10(0.0815) = 1.09
Now with the ten-fold dilution
[H+] = √((5.9x10-2 x ) / 2) = mol dm-3
pH = -log10[H+] = -log10(0.0257) = 1.59….i.e. ten-fold dilution increased pH by 0.50

10. Example 3: Calculating Kb from pH
A mol dm-3 solution of methylamine (CH3NH2) has a pH of Determine Kb of methylamine and Ka of the methylammonium ion (CH3NH3+).
Since pH = 11.59
pOH = 14 – = 2.41
[OH-] = = 3.92x10-3 mol dm-3
Remember:
[BOH] at equilibrium is same as stated in question
[OH-] = [B+]
Kb = [B+][OH-]/[BOH] Known values subbed in
Kb = (3.92x10-3).(3.92x10-3)/0.0350
Kb = 4.39x10-4
To calculate Ka of the conjugate acid use:
Ka x Kb = Kw
Ka = Kw / Kb
= 1.00x10-14 / 4.39x10-4
= 2.78x10-11

11. Example 4: Calculating Ka, pKa, Kb and pKb from each other
The pKa of benzoic acid is Calculate Ka, Kb and pKb
Ka = 10-pKa = = 6.31x10-5
Kb x Ka = Kw
Kb = Kw / Ka = 1.00x10-14 / 6.31x10-5 = 1.58x10-10
pKb = -log10(Kb) = -log10(1.58x10-10) = 9.80 OR
Since pKa + pKb = pKw
pKb = 14 – pKa = 14 – 4.20 = 9.80

12. Practice Time Work through the calculations on the worksheet
If you finish early, complete a flow-chart that can be used to answer weak acid/base questions

13. Key Points Calculations rely on two key assumptions:
Concentration of HA or BOH at equlibrium is the same as given in the question
Reasonable as equilibrium effects mean dissociation is often 1% or less
[H+]/[OH-] and [A-]/[B+] are not separate variables but are related to each other
Expressing [A-]/[B+] in terms of [H+]/OH-] is a key step