Acids, bases and Ph + Neutralisation and buffers

Question 1

what do we now call Acids and Bases

Answer 1

Brønsted-lowry acid/base

Question 2

Mono- di- tri-basic acids

Answer 2

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

Question 3

Half equation for the basic acid reactions

Answer 3

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^+

Question 4

Weak vs strong acid

Answer 4

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+

Question 5

Acid + water

Answer 5

Acid + H20 —-> H3O^+ (oxonium) + metal ion

Water here acids as a base + accepts the proton to produce a hydrated hydrogen ion

Question 6

Conjugate acid base pairs

Answer 6

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)

Question 7

Ka - acid dissociation constant

Answer 7

this is the extent of acid dissociation (in weak acids + homo equ)

equ= Ka= [H+] [A-] / [HA]

units = moldm -3

Question 8

Significance of Ka

Answer 8

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

Question 9

pKa equation + reason for use

Answer 9

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)

Question 10

Converting between Pka and Ka

Answer 10

Ka = 10^-pKa

pKa = - log Ka

Question 11

Kw- ionic product of water
equation + explanation

Answer 11

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)

Question 12

Importance of Kw

Answer 12

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)

Question 13

Importance of Kw

Answer 13

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)

Question 14

PH equation

Answer 14

PH = -log [H+]

low PH = High [H+]
and reverse

PH= logarithmic so every 1 decrease is 10 × greater H+ conc

Question 15

convertion between PH and H+

Answer 15

PH= -log [H+]

[H+] = 10 ^-PH

round conc to no of sig fig that OH has d.p

Question 16

convertion between PH and H+

Answer 16

PH= -log [H+]

[H+] = 10 ^-PH

round conc to no of sig fig that PH has d.p

Question 17

Find PH of strong monobasic acid

Answer 17

HA —> H+ + A-
completely dissociates

so for strong [HA] = [H+]
as 1:1 ratio

so then just do
PH = -log [H+]

Question 18

finding PH of weak monobasic acid

Answer 18

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+]

Question 19

finding PH of weak monobasic acid LIMITATIONS

Answer 19

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-]

Question 20

Finding PH of a strong Base

Answer 20

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+]

Question 21

Buffer definition

Answer 21

mixture that minimises PH changes on addition of SMALL amounts of acid or base

Question 22

what is a buffer solution made up of?

Answer 22

weak acid (HA)

conjugate base (A-)

Question 23

basic buffer info

Answer 23

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

Question 24

2 ways to prep a buffer

Answer 24

1) mix weak acid and 1 of its salts

2)react EXCESS weak acid with strong Alkali

Question 25

How are reservoirs of acid and conjugate ions formed

Answer 25

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

Question 26

How doe buffers work on addition of small amounts of acid?

Answer 26

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

Question 27

How do buffers work on addition of small amounts of base

Answer 27

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

Question 28

Exam tips for Buffer Q

Answer 28

state:
relevant equ equation
define buffer
how it minimises effect on PH for acid + base

Question 29

How to calc PH of a buffer

Answer 29

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

Question 30

How to calc ratio of a buffer

Answer 30

Ka = [H+][A-]
[HA]

[H+] = 10^-PH

then:
[Ka]/[H+] = [A-]/[HA]
treat [HA] as 1 to find ratio of [HA] : [A-]

Question 31

Control of blood PH

Answer 31

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