O - Acids, pH, pKa (pH and pKa)

Question 1

what is the conc of [H+]?

Answer 1

[H+] = 1 x10-4 moldm-3 = pH4

Question 2

how do you work out pH?

Answer 2

-log [H+(aq)]

Question 3

how do you work out the conc. of strong acids?

Answer 3

[HA] = [H+]

because it fully dissociates

Question 4

what is the pH of a 0.01moldm-3 solution of a strong acid?

Answer 4

[H+] = 0.01
pH = -log(0.01)
pH = 2

Question 5

how do you calculate the value of [H+] from pH?

Answer 5

10-ph
e.g HA has a pH of 2.6 what is the [H+]
[H+] = 10-2.6
= 2.5x10-3

Question 6

assumption 1:

Answer 6

[A-] = [H+]

means we ignore the dissociation of water

Question 7

assumption 2:

Answer 7

[HA] initial = [HA] equilibrium

H+ must come from somewhere but we ignore it because the eqm is way to the left

Question 8

find pH of a solution of ethanoic acid with a conc. of 1 mol dm-3 = ka of ethanoic acid 1.7x10-5 moldm-3 at 298k

using assumption 1 and 2

Answer 8

HA —> H+ + A-
expession: Ka = [H+][A-]/[HA]
rearrange for [H+]: [H+] =√[HA]Ka
assumption 2 - [HA] = 1
[H+] = 4.12 x10-3 moldm-3
pH = -log(ans)
ph = 2.38

Question 9

what is pKa?

Answer 9

weak acids have a very low Ka so they use the log pKa

-log(Ka)

Question 10

what is the ionic product of water?

Answer 10

Kw = 1x10-14 mol2dm-6

Question 11

ionic product of water = Kw

write a Kw expression for

H2O <=> H+ + OH-

Answer 11

Ka = [H+][OH-]
this then becomes
Kw = [H+][OH-]
it is this that is 1x10-14