Transcript Weak Acid and Base Calculations

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?
A.
B.
C.
D.
Acid
HA
HB
HC
HD
pKa
2.0
–
4.0
–
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Ka
–
1 × 10–3
–
1 × 10–5
Recap
WEAK ACIDS
WEAK BASES
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Ka = [H+][A-]/[HA]
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Kb = [B+][OH-]/[BOH]

pKa = -log10(Ka)

pKb = -log10(Kb)

Smaller Ka  weaker acid

Smaller Kb  weaker base

Smaller pKa  stronger acid

Smaller pKb  stronger base
Acid
Ka
pKa
Base
1.00
0.00
Oxalic acid, HO2CCO2H
5.9x10-2
Hydrofluoric, HF
Kb
pKb
Diethylamine, (C2H5)2NH
1.3x10-3
2.89
1.23
Ethylamine, C2H5NH2
5.6x10-4
3.25
7.2x10-4
3.14
Methylamine, CH3NH2
4.4x10-4
3.36
Methanoic, CHOOH
1.77x10-4
3.75
Ammonia, NH3
1.8x10-5
4.74
Ethanoic, CH3COOH
1.76x10-5
4.75
Phenol, C6H5OH
1.6x10-10
9.80
Hydronium ion, H3O+
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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?
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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:
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Ka × Kb = Kw = 1.00x10-14

pKa + pKb = pKw = 14.0

pH + pOH = pKw = 14.0
And from previous lessons:
[ B  ][OH  ]
Kb 
[ BOH ]
[ H  ][ A ]
Ka 
[ HA]

pKa = -log10(Ka)
pKb = -log10(Kb)

pH = -log10[H+]
pOH = -log10[OH-]
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Calculating the pH of a weak acid 1
6 of 51
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Calculating the pH of a weak acid 2
7 of 51
© Boardworks Ltd 2010
Calculations of weak acid pH
8 of 51
© Boardworks Ltd 2010
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:



= [NH4+][OH-]/[NH3]
= [NH4+][OH-]/ 0.500
Sub all known values into equation
Looks like there are 2 unknowns
However, since [NH4+] = [OH-]:
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Kb
1.8x10-5
1.8x10-5 = [OH-]2 / 0.500
[OH-] = √(1.8x10-5 x 0.500)
[OH-] = 0.0030 mol dm-3
Rearrange to make [OH-] the subject
Perform calculation
% 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
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Sub in all known values
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
[H+] = √((5.9x10-2 x 0.225) / 2) = 0.0815 mol dm-3

pH = -log10[H+] = -log10(0.0815) = 1.09


Ka = [H+][A-] / [HA]
5.9x10-2 = [H+][A-] / 0.225
Rearrange to make [H+] subject
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
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Remember:
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To calculate Ka of the conjugate acid use:
Ka x Kb = Kw
Ka = Kw / Kb
= 1.00x10-14 / 4.39x10-4
= 2.78x10-11
[BOH] at equilibrium is same as stated in question
[OH-] = [B+]
Kb = [B+][OH-]/[BOH]
Kb = (3.92x10-3).(3.92x10-3)/0.0350
Kb = 4.39x10-4
Known values subbed in
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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
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OR
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Since pKa + pKb = pKw
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pKb = 14 – pKa = 14 – 4.20 = 9.80
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Practice Time

Work through the calculations found here.
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If you finish early, complete a flow-chart that can be used to answer weak
acid/base questions
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Key Points
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Calculations rely on two key assumptions:
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Concentration of HA or BOH at equilibrium is the same as given in the question
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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
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Expressing [A-]/[B+] in terms of [H+]/OH-] is a key step
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