Acids, bases and Ph + Neutralisation and buffers Flashcards
what do we now call Acids and Bases
Brønsted-lowry acid/base
Mono- di- tri-basic acids
Mono- releases 1 proton per molecule
+ requires 1 mole of OH- to neut
di- 2 protons + 2 moles to neutralize
tri- 3 protons + 3 moles to neutralize
Half equation for the basic acid reactions
Metal + acid
e.g.
Mg + 2H^+ —> Mg^+2 + H2
Metal oxide + acid
e.g.
O^-2 + 2H^+ —–> H2O
Alkali + acid
e.g.
OH^- + H^+ —-> H2O
Carbonate + acid
e.g.
CO3^2- + 2H^+ —-> CO2 + H2O
ammonia + acid
e.g.
NH3 + H^+ —-> NH4^+
Weak vs strong acid
weak-
partially dissociates into its ions in solution
Equ pos is far left, so reverse reaction is sig (partial) —> conc of H+ is low compared
strong-
completely dissociates into its ions in solution
In a strong acid equ pos is so far right reverse is insignificant (complete) —> high conc of H+
Acid + water
Acid + H20 —-> H3O^+ (oxonium) + metal ion
Water here acids as a base + accepts the proton to produce a hydrated hydrogen ion
Conjugate acid base pairs
this is a pair of 2 species that transform into eachother by the gain or loss of a proton
only in partial dissociation
e.g.
HCl + H20 <—> H30^+ + Cl-
A1 B2 A2 B1
in forward reaction HCl = acid and Cl- = base it forms (conjugate acid base pair 1)
In backward reaction H30+ = acid and H20 = base it forms (con acid base pair 2)
Ka - acid dissociation constant
this is the extent of acid dissociation (in weak acids + homo equ)
equ= Ka= [H+] [A-] / [HA]
units = moldm -3
Significance of Ka
Ka = large
equ pos is far to the right
lots of HA has dissociated into its ions (suggests strong)
Ka = small
equ pos far left
a lot og HA is not dissociated into its ions (suggests weak)
Largest Ka = strongest acid
ONCE AGAIN ONLY EFFECTED BY TEMP
pKa equation + reason for use
pKa = -log Ka
Ka if often very large/ small and difficult to use, this way it is more manageable as it is logarithmic
(p = -log)
Converting between Pka and Ka
Ka = 10^-pKa
pKa = - log Ka
Kw- ionic product of water
equation + explanation
Pure water can self dissociate/ self ionise
H2O <—> H+ + OH-
H2O + H2O <—-> H3O+ + OH-
very small extent of dissociation (1 per 500 million)
Kw = [H+] [OH-]
units = mol2 dm-6
at 25⁰C Kw = 1 x 10^-14 mol2 dm-6
(H2O not in equation as basically stable)
Importance of Kw
in all aqueous solutions H+ and OH- are present
so
[H+] [OH-] = Kw
acidic = H+ > OH-
Neutral = H+ = OH-
Basic = H+ < OH-
Can be used to determine conc of H+ and OH -
[H+] = Kw/[OH-]
[OH-] = Kw/[OH-]
+ to prove solution is neutral
as in neut
Kw = [H+]² or [OH-]²
do just square root Kw
then PH = -log ans = 7
ONLY EFFECTED BY TEMP
(SELF DISSOCIATION = ENDO)
Importance of Kw
in all aqueous solutions H+ and OH- are present
so
[H+] [OH-] = Kw
acidic = H+ > OH-
Neutral = H+ = OH-
Basic = H+ < OH-
Can be used to determine conc of H+ and OH -
[H+] = Kw/[OH-]
[OH-] = Kw/[OH-]
+ to prove solution is neutral
as in neut
Kw = [H+]² or [OH-]²
do just square root Kw
then PH = -log ans = 7
ONLY EFFECTED BY TEMP
(SELF DISSOCIATION = ENDO)
PH equation
PH = -log [H+]
low PH = High [H+]
and reverse
PH= logarithmic so every 1 decrease is 10 × greater H+ conc
convertion between PH and H+
PH= -log [H+]
[H+] = 10 ^-PH
round conc to no of sig fig that OH has d.p
convertion between PH and H+
PH= -log [H+]
[H+] = 10 ^-PH
round conc to no of sig fig that PH has d.p
Find PH of strong monobasic acid
HA —> H+ + A-
completely dissociates
so for strong [HA] = [H+]
as 1:1 ratio
so then just do
PH = -log [H+]
finding PH of weak monobasic acid
Only partially dissociates so can’t assume HA = H+
make assumptions:
large proportion of HA doesn’t dissociate
so
[HA] equ ~ [HA] undis
also
negligible dissociation of water means
assume [H+] ~ [A-]
So use Ka = [A-] [H+] / [HA]
assume A- ~ H+
so Ka = [H+]² / [HA]
assume HA equ ~ HA undis
so square root (Ka × [HA] ) = [H+]
then PH = -log [H+]
finding PH of weak monobasic acid LIMITATIONS
Stronger weak acids have larger Ka and greater dissociation
so can’t assume [HA] equ ~ [HA] undis
weaker weak acids/ very dilute will have very low [H+]
so [H+] from self Diss of water will be sig
so can’t assume [H+] ~ [A-]
Finding PH of a strong Base
Strong base= Alkali that comp dissociates into OH-
So you can use ratio of Base:OH-
to calc conc of OH-
then Kw= [H+][OH-]
then Kw/[OH-] = [H+]
then PH= -log [H+]
Buffer definition
mixture that minimises PH changes on addition of SMALL amounts of acid or base
what is a buffer solution made up of?
weak acid (HA)
conjugate base (A-)
basic buffer info
Can have specific PH
Ph less than 7 = acid buffer (only one we have to know)
the weak acid with the closest pKa to desired Ph is chosen
All reactants and products are AQUEOUS
2 ways to prep a buffer
1) mix weak acid and 1 of its salts
2)react EXCESS weak acid with strong Alkali
How are reservoirs of acid and conjugate ions formed
weak acid partially dissociates:
e.g. CH3COOH <–> CH3COO- + H+
large small small
Salt of weak acid completely dissociates:
e.g. CH3COONa –> CH300- + NA+
small large large
Overall:
e.g. CH3COOH <–> CH3COO- + H+
large large small
NOTE- add of salt increases conc of A- pushes equ of HA further left but shift = minimal so doesn’t effect HA conc
How doe buffers work on addition of small amounts of acid?
conjugate base reacts with H+ ion
e.g. CH3COO- + H+ —> CH3COOH
to produce acid
Equilibrium shifts left to minimise the effect of an increase in [H+]
e.g. CH3COOH <–> CH3COO- + H+
Most of the added H+ ions are removed
How do buffers work on addition of small amounts of base
H+ ions in equ react with OH- to form H20
Causing acid to further dissociate and equ pos to shift right to minimise effect
e.g. CH3COOH <–> CH3COO- + H+
Most of the H+ ions are restored
Exam tips for Buffer Q
state:
relevant equ equation
define buffer
how it minimises effect on PH for acid + base
How to calc PH of a buffer
in a buffer [H+] doesn’t = [A-]
-log ((Ka * [HA]) / [A-])
because- PH= -log([H+])
and [H+]= ((Ka * [HA])/ A- ) in this case
OR Henderson Hasselbach equ:
PH = pKa + log { [A-]/[HA] }
If weak acid + salt then use this
if weak acid + strong alkali then use rice table then this
How to calc ratio of a buffer
Ka = [H+][A-]
[HA]
[H+] = 10^-PH
then:
[Ka]/[H+] = [A-]/[HA]
treat [HA] as 1 to find ratio of [HA] : [A-]
Control of blood PH
Ph of blood controlled by multiple buffer ( to keep between 7.35-7.45)
Carbonic acid Hydrogen carbonate buffer system:
H2CO3(Aq)<–>H+(Aq) + HCO3-(Aq)
weak acid conjugate base
Works same as other buffers
Carbonic acid is then converted to CO2 in lungs and exhaled
