Save
Chemistry Paper 1
Acid, bases and buffers
Save
Share
Learn
Content
Leaderboard
Learn
Created by
Cailyn
Visit profile
Cards (76)
Kw
ionic
product of
water
Monobasic acid
Acid that donates
one
mole of a
proton
Brønsted-Lowry acid
Species that
donates
a
proton
monobasic acids
HCl
HCN
HNO3
CH3COOH
dibasic acid
H2SO4
H2CO3
tribasic acids
H3PO4
H3BO3
conjugate acid-base pair
pair of two species that can be
interconverted
by a
transfer
of
protons
A1 B2 —> A2 B1
acids have a
positive
charge
bases have a
negative
charge
larger
Ka means more
dissociation
and a stronger acid
Ka
acid
dissociation
constant
Ka of weak acids
only
paritally
dissociate
so theres more molecules of weak acid than of H+ and A-
Assumption of Ka of weak acids
H+ = A-
HA dissociates to produce
equal
concentrations of the two
concentrant on of H+ due to the ionisation of
water
is
negligible
compared to the H+ concentration of the acid
Brønsted-Lowry base
species that
accepts
a proton
Dibasic acid
one
mole of an acid can donate/release
2
protons
tribasic acid
one
mole of an acid can release/ donate
3
protons
acids
dissociate and release H+ ions in
aqueous
solutions and
alkalis
dissociate and release OH- in
aqueous
solution
Calculating pH of strong Monobasic acids
pH = -log[
H+
]
calculating the pH of strong bases
Kw = [OH-][H+]
calculating the pH of
weak
acids
Ka= [H+]²/[HA]
Kw
increases
with increasing temperature as
more
bonds are broken (
endothermic
process)
assumptions for calculating pH of weak acids
[HA] at the start is the same as [HA] in equilibrium
remains
constant
as
dissociation
is small
this doesn’t apply when there’s a
stronger
weak acid
assumption for pH of weak acids do not always work
-[H+] doesnt equal [A-] if the acid is very
dilute
-[HA] is not the same throughout if there’s a
stronger
weak acid
range of weak acids lies between pH
3
and
6
pH of strong monobasic acids
complete
dissociation in solution
[
H
+] = [
HA
]
pH of strong acid can be calculated directly from the
concentration
of an acid
Kw
if [H+(aq)] > [OH- (aq)] then the solution is
acidic
Kw
if [H+(aq)] < [OH- (aq)] then the solution is
alkaline
pH of a
buffer
solution
[H+] = Ka x [HA] / [A-]
what is a buffer
solution?
solution that
minimises
pH
changes
on the addition of small amounts of
bases
or
acids
buffer
[H+] / Ka = [HA] / [A-]
ratio of
acid
to its
conjugate
base
Kw
if temperature
increases
, acid
dissociates
to a greater extent and so the Kw is higher
acid + base —>
salt
+
water
acid + metal oxide —>
salt
+
hydrogen
(gas)
acid + alkali —>
salt
+
water
How are buffers formed?
weak
acids and the
salt
of a weak acid
weak acid
partially
dissociates in solution and forms a relatively
low
concentration of A- ion
salt
fully
ionises in solution
reserve supplies of acid and its conjugate base
formation of buffers example
weak acid
and salt of weak acid
ethanoic
acid
and
sodium
ethanoate
CH3COOH / CH3COONa
How are buffers formed?
partial
neutralisation
of the weak acid
add aq alkali to
excess
weak acid
weak
acid is
partially
neutralised by the
alkali
forming the conjugate base
some of the
weak
acid is left over unreacted so a mixture of
salt
and
unreacted
acid forms
buffer formation
partial neutralisation e.g CH3COOH
CH3COOH + NaOH —>
CH3COONa
+
H2O
CH3COOH
⇌
CH3COO-
+
H+
action of a buffer
Addition of an
acid
:
equilibrium shifts to the
left
added H+ react with
conjugate
base
[H+]
increases
and pH is
lowered
action of a buffer
Addition of an alkali:
equilibrium shifts to the
right
(to replenish lost H+)
H+ (aq) + OH- (aq) —>
H2O
(l)
HA
dissociates
[OH-]
increases
and pH increaases
buffer is most effective when there are
equal
concentrations of
weak
acid and its
conjugate
base
calculation of the pH of a buffer when its partial neutralisation
n(acid used)
n(alkali) = n(conjugate base formed)
n(acid
remaining)
= n(acid
used)
- n(
alkali
added)
See all 76 cards