Transcript Weak acid and base calculations

Weak acid and base
calculations
What’s so hard?
Unlike strong acids and bases, weak examples do not dissociate fully.
For example, a 1 molL-1 HCl solution dissociates fully, so we make the
approximation that the [H3O+] = 1 molL-1
For a weak acid e.g. CH3COOH, dissociation is only partial, so we cannot
make the same approximation.
Example
Calculate the pH of a 0.1 molL-1 ethanoic acid
solution. Ka = 1.74 x 10-5
Step 1: Write a Ka expression for the dissociation of the weak acid.
CH3COOH + H2O  CH3COO- + H3O+
Ka = [H3O+][CH3COO-]
[CH3COOH]
Step 2: Make 2 assumptions
1. That the amount of dissociation is so small it is insignificant. Therefore
[CH3COOH] = 0.1 molL-1 in this example
2. That [H3O+] = [CH3COO-] (we will call this X)
Example ctd…
Step 3: Substitute the values in and solve for X
Ka = [H3O+][CH3COO-]
[CH3COOH]
1.74 x 10-5 =
[X][X]
[0.1 molL-1]
X2 = 1.74 x 10-5 x 0.1 molL-1
X = √ 1.74 x 10-5 x 0.1 molL-1
[H3O+] = 1.32 x 10-3 molL-1
Step 4: Calculate the pH in the usual way
pH = -log [H3O+]
= -log (1.32 x 10-3)
= 2.88
Now try these:
1. Calculate the pH of an 0.1 molL-1 propanoic acid solution given Ka
(propanoic acid) = 1.4 x 10-8
Answer = 4.43
2. Calculate the pH of an 0.02 molL-1 ammonium chloride solution given pKa
(ammonium ion) = 9.24
Answer = 5.46
3. Calculate the pH of a 1 x 10-3 molL-1 hydrogen fluoride solution given Ka
(HF) = 6.76 x 10-4
Answer = 3.09
Now try a weak base!
Calculate the pH of a 0.25 molL-1 ammonia solution.
Ka (ammonium ion) = 1 x 10-9
1. For weak base calculations, Kb will never be given so you will have to
calculate it
Ka x Kb = 1 x 10 -14
1 x 10-14 / 1 x 10-9 =
1 x 10-5
2. Write a Kb expression as before
NH3 + H2O  NH4+ + OHKb = [OH-][NH4+]
[NH4]
Weak base ctd……
Step 2: Make 2 assumptions
1. That the amount of dissociation is so small it is insignificant. Therefore
[NH3] = 0.25 molL-1 in this example
2. That [OH-] = [NH4+] (we will call this X)
Step 3: Substitute the values in and solve for X
Kb = [OH-][NH4+]
[NH3]
1 x 10-5 = [X][X]
[0.25 molL-1]
X2 = 1 x 10-5 x 0.25 molL-1
X
[OH-]
= √ 1 x 10-5 x 0.25 molL-1
= 1.58 x 10-3 molL-1
Weak base ctd….
Step 4: Calculate the pH in the usual way
To calculate H3O+ we need to use pH + pOH = 14 (or [OH-][H3O+]=1 x 10-14)
pOH = - log [OH-]
pH = 14 – pOH
pH = 14 – 2.80
=
11.2
Now try these!
1. Calculate the pH of an 0.55 molL-1 solution of sodium ethanoate, given
the Ka for ethanoic acid = 1.74 x 10-5