Title: Lesson 8: Weak acid/base calculations

Learning Objectives:– Perform calculations involving weak acids

– Perform calculations involving weak bases

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Based on information in the table below, which acid is the strongest?

Acid pKa Ka

A. HA 2.0 – B. HB – 1 × 10–3

C. HC 4.0 – D. HD – 1 × 10–5

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RecapWEAK ACIDS

Ka = [H+][A-]/[HA]

pKa = -log10(Ka)

Smaller Ka weaker acid

Smaller pKa stronger acid

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

WEAK BASES

Kb = [B+][OH-]/[BOH]

pKb = -log10(Kb)

Smaller Kb weaker base

Smaller pKb stronger base

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

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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 0.250 M solution of weak acid Z is 5.4, what is it’s Ka and pKa?

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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-]

][

]][[

BOH

OHBKb

][

]][[

HA

AHKa

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Calculating the pH of a weak acid 1

7 of 51 © Boardworks Ltd 2010

Calculating the pH of a weak acid 2

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Calculations of weak acid pH

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Example 1: Calculation of [OH-] What is the concentration of OH- ions in a 0.500 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-]/ 0.500 Looks like there are 2 unknowns

However, since [NH4+] = [OH-]:

1.8x10-5 = [OH-]2 / 0.500 Rearrange to make [OH-] the subject [OH-] = √(1.8x10-5 x 0.500) Perform calculation [OH-] = 0.0030 mol dm-3

% Dissociation = 0.0030 / 0.500 x 100 = 0.60%

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Example 2: Calculating pH What is the pH of a 0.225 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-] / 0.225 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 / 0.225 Rearrange to make [H+] subject [H+] = √((5.9x10-2 x 0.225) / 2) = 0.0815 mol dm-3

pH = -log10[H+] = -log10(0.0815) = 1.09

Now with the ten-fold dilution [H+] = √((5.9x10-2 x 0.0225) / 2) = 0.0257 mol dm-3

pH = -log10[H+] = -log10(0.0257) = 1.59….i.e. ten-fold dilution increased pH by 0.50

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Example 3: Calculating Kb from pH A 0.0350 mol dm-3 solution of methylamine (CH3NH2) has a pH of 11.59. Determine Kb

of methylamine and Ka of the methylammonium ion (CH3NH3+).

Since pH = 11.59 pOH = 14 – 11.59 = 2.41

[OH-] = 10-2.41 = 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

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Example 4: Calculating Ka, pKa, Kb and pKb from each other

The pKa of benzoic acid is 4.20. Calculate Ka, Kb and pKb

Ka = 10-pKa = 10-4.20 = 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

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Practice Time

Work through the calculations found here.

If you finish early, complete a flow-chart that can be used to answer weak acid/base questions

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Key Points

Calculations rely on two key assumptions: Concentration of HA or BOH at equilibrium 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