Acid-base Equilibria

Question 1

Brønsted–Lowry acid

Answer 1

proton donor

Question 2

Brønsted–Lowry base

Answer 2

proton acceptor

Question 3

acid-base reactions involve the transfer of

Answer 3

protons

Question 4

pH

Answer 4

-log10[H+]

Question 5

[H+] from pH

Answer 5

10^-ph

Question 6

strong acid

Answer 6

fully dissociated

Question 7

weak acid

Answer 7

partially dissociated

Question 8

pH of a strong acid

Answer 8

no ka

[H+]=[HA]

Question 9

pH of a weak acid

Answer 9

ka=[H+]^2/[HA]

Question 10

assumptions of a weak acid

Answer 10

1. [A-]=[H+] as no H+ from the dissociation of water only from the acid
2. [HA] equilibrium = [HA] original as dissociation is negligable

Question 11

Kw

Answer 11

[H+][OH-]=[H+]^2
=1x10^-14
so pH of water is 7

Question 12

pH of strong base

Answer 12

using kw

[H+]=kw/[OH-]

Question 13

pKa

Answer 13

-log10(Ka)

Question 14

pKw

Answer 14

-log10(Kw)

Question 15

dilution of strong acids

Answer 15

for each dilution of x10 pH increases by 1

Question 16

dilution of weak acids

Answer 16

Weak acids are in equilibrium
as they are diluted
some of the undissociated acid molecules split up
pH does not increase as fast as strong acid
each dilution of 10x, the pH increases by 0.5 unit, and for each dilution of 100x, the pH increases by 1 unit

Question 17

numbers for pH curves

Answer 17

13 11
8 7 6
3 1

Question 18

how to select a suitable indicator

Answer 18

needs to change colour in the vertical region,
pH changes sharply so accurate
pH range within vertical region

Question 19

end point

Answer 19

point when the indicator changes colour

should be in the middle of the vertical section of the line

Question 20

buffer solution

Answer 20

resists the changes in pH if you add a small amount of acid or base

Question 21

how does a buffer solution work

Answer 21

contains high of both [HA] and [A-]
adding small H+: H+ + A- > HA
adding small OH-: HA + OH- > A- + H2O
[HA] : [A-] stays aprox constant
[H+] and pH stays aprox constant

Question 22

pH of a buffer

Answer 22

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

Question 23

at 1/2 equivalence

Answer 23

pKa = pH as
[HA]=[A-] so
Ka=[H+] as they cancel out

Question 24

expression for Kin / Ka of indicator

Answer 24

=[H+][in-]/[Hin]

Question 25

at end point, Kin

Answer 25

[in-]=[Hin] so 
[in-]/[Hin]=1 and
[H+] = Kin
-log[H+]=-log[Kin]
pH=pKin

Question 26

difference in enthalpy changes of neutralisation values for strong and weak acids

Answer 26

strong acids: HA>H+ + A- all H+ the add alkali to neutralise
weak acids: HA <>H+ + A- add alkali to neutralise
equi shifts right to replace H+
HA dissociation is endothermic
some energy is reabsorbed so enthalpy is less -ve

Question 27

pH of blood

Answer 27

7.4

Question 28

buffers of the blood

Answer 28

carbonate/hydrogencarbonate base pairs
H2CO3(aq) >HCO3-(aq) + H+(aq) to decrease H+
H2CO3(aq) >CO2(aq) + H2O(l) to increase H+