Topic 18: Acids and bases
18.1 Lewis acids and bases
18.2 Calculations involving acids and bases
18.3 pH curves
Kw
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•
•
•
Ionic product constant of water is an equilibrium constant
Affected by temperature
Dissociation of water is endothermic
Higher temperature = greater dissociation = higher [H+] (and [OH-])= lower
pH
• The pH doesn’t actually tell us what is neutral! Water dissociates more at
higher temperatures (lower pH) but the concentrations of H+ and OH- are
equal
– pH 7 is applicable for 298 K only
• Always state temperature with pH measurements
pOH
• Subtract pH from 14
• pOH = -log10[OH-]
pKw
• An expression that relates Kw to pH and pOH
• pKw = log10(Kw) or Kw = 10-pKw
• pKw = pH + pOH at any temperature
– i.e. at 298 K, pKw = 14
Strong acids and bases
• Recall SL
• Assume full dissociation
pH & pOH can be calculated directly
from concentration of the solution
• E.g. 0.20 mol dm-3 NaOH would
yield a 0.20 mol dm-3 OH- solution
• pOH = -log10[OH-] = 0.7
• pH = 14 – pOH = 13.3
Weak acids and bases
• Incomplete dissociation = can’t use
concentration of solutions
• Need to know the extent of the dissociation, i.e.
where the position of equilibrium lies
Kc =
[H3O+][A-]
[HA][H2O]
conjugate base
acid
OR
• write an expression specifically for this topic
that takes out water, since water is a constant:
[H3O+][A-]
Ka =
[HA]
• Ka is called the acid dissociation constant and
has a fixed value for a particular acid at a
specified temperature
Weak acids and bases
• The higher the Ka value, the stronger
the acid
• Since derived from an equilibrium
expression, the only factor that
affects the value of Ka for a particular
acid is temperature (i.e. not
concentration)
– This is a useful property when we want
to limit (buffer) changes in the pH of a
system when ion concentrations
change: Le Chatelier will ‘correct’ to
maintain Ka… more on this in 18.3…
• We can use the same type of
equation for bases:
• Kb = base dissociation constant
Ka =
[H3O+][A-]
[HA]
conjugate acid
[BH+][OH-]
Kb =
[B]
base
Calculations using Ka and Kb
We can use known concentrations of an acid or base + its experimentally
derived pH value (e.g. with a probe) to calculate Ka or Kb
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•
•
•
10–pH tells us [H3O+] at equilibrium
OR
at 298 K, 14 – pH gives us pOH, so 10pH–14 tells us [OH-] at equilibrium
the concentration of conjugate acid/base can be determined from its formula, e.g. if we know
[H3O+] for ethanoic acid, we know [CH3COO-]
the concentration of the undissociated acid or base can be found by subtracting the concentration
of its conjugate from the original concentration given
– For weak acids and bases, this goes past the significant figures in the original value (since the
dissociation is very small), so you may not actually need to change the acid/base concentration
in this case, i.e. use the original concentration as the equilibrium concentration
Alternatively, we might know the Ka or Kb value for a give acid or base and then we can work
out the resulting pH for any specified concentration we want to use.
E.g. Calculate the pH of a 1.5 x 10-2 mol dm-3
solution of ethanoic acid at 25 C, given that
we know its Ka value is 1.74 x 10-5
answer on
next slide…
Calculations using Ka and Kb
E.g. Calculate the pH of a 1.5 x 10-2 mol dm-3 solution of ethanoic acid at 25 C, given that
we know its Ka value is 1.74 x 10-5.
Ka =
[H3O+] [CH3COO-]
[CH3COOH]
[H3O+]2
1.74 x 10-5 = [CH COOH]
3
[H3O+]2
1.74 x 10-5 = 1.5 x 10-2
2.63 x 10-7 = [H3O+]2
5.12 x 10-4 = [H3O+]
pH = -log10[H3O+]
pH = 3.29
or
[H3O+]2
[CH3COOH]
since [H3O+] = [CH3COO-]
here we have to assume that
for a very small K value, we can
let the concentration of acid
remain unchanged
pKa and pKb
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•
Simply alternative ways of presenting values to fit into a more user-friendly scale
Identical to converting [H3O+] or [OH-] into pH and pOH
same as pH: logarithmic
-pKa
i.e. pKa = -log10Ka /
Ka = 10
scale (one unit change in pKb
pKb = -log10Kb /
Kb = 10-pKb
is a 10-fold change in Kb etc.)
The values in Table21 of your Data Booklet are all reported as pKa or pKb.
pKa and pKb
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•
pKa + pKb = pKw
pKa + pKb = 14.00 at 298 K
Why? Let’s see how this works out mathematically by writing expressions for Ka and Kb
with an imaginary acid, HA, and it’s conjugate base, A-.
regenerating the acid is
the conjugate…of the
Ka = [H3O+][A-]
Kb = [HA][OH-]
x
conjugate!
[HA]
[A-]
What happens if we multiply these expressions together?
Go through and cross cancel…
conjugate base from the
Ka expression
We are left with Ka x Kb = [H3O+] x [OH-] and we already know that [H3O+][OH-] is the
expression for Kw.
Ka x Kb = Kw
or taking the log of everything we get
(note the change to plus now in logarithmic form)
pKa + pKb = pKw
pKa and pKb
•
•
pKa + pKb = pKw
pKa + pKb = 14.00 at 298 K
This also means we can quantitatively discuss why strong acids have weak conjugate
bases and vice versa, since the concentrations of the hydronium and hydroxide ions they
produce still reach the same equilibrium constant for water (Kw).
Exam Q
Exam Q
Exam Q
Recall that Ka = 10–pKa
Need to do approximate calculations
(no calculators in Paper 1)
Exam Q
Same….but different!