pH and buffers

Question 1

conjugate acid-base pair

Answer 1

a pair of two species that transform into each other by the gain or loss of a proton (H+)

Question 2

acid dissociation constant equation

Answer 2

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

Question 3

significance of the values of Ka

Answer 3

Ka > 1
equilibrium is far right
a lot of the acid is dissociated into its ions, it is a strong acid
Ka < 1
equilibrium is far left
a lot of the acid is not dissociated, it is a weak acid

Question 4

conversion between Ka and pKa

Answer 4

pKa = -log Ka
Ka = 10^-pKa

Question 5

ionic product of water equation

Answer 5

Kw = [H+][OH-]

Question 6

conversion between [H+] and pH

Answer 6

pH = -log [H+]
[H+] = 10^-pH

Question 7

significance of Kw

Answer 7

[H+] > [OH-] acidic
[H+] < [OH-] alkaline
[H+] = [OH-] neutral

Question 8

find the pH of a solution of hydrochloric acid of concentration 0.10moldm-3

Answer 8

HCl(aq) is a strong acid, so complete dissociation:
HCl(aq) → H+ (aq) + Cl- (aq)
0.10 moldm-3 → 0.10 moldm-3
pH = -log [H+]
pH = -log 0.10
pH = 1.00

Question 9

find the pH of a solution of butanoic acid of concentration 0.01moldm-3
(the Ka value for butanoic acid is 1.51 x 10-5moldm-3)

Answer 9

C3H7COOH (aq) → C3H7COO- (aq) + H+ (aq)
Ka = [C3H7COO-] [H+] / [C3H7COOH]
at equilibrium: [C3H7COO-] ~ [H+]
Ka = [H+]^2 / [C3H7COOH]
sub in values: 1.51 x 10^-5 = [H+]^2 / 1.00 x 10^-2
[H+]^2 = 1.51 x 10^-7
[H+] = 3.89 x 10^-4 mol dm-3
pH = -log[H+] = -log 3.89 x 10^-4 = 3.411

Question 10

find the pH of a solution of sodium hydroxide of concentration 0.010moldm-3 at 25°C

Answer 10

NaOH (aq) → Na+ (aq) + OH- (aq)
0.010 moldm-3 → 0.010 moldm-3
Kw = [H+] [OH-]
1.00 x 10^-14 = [H+] x 0.010
1.00 x 10^-14 / 0.010 = [H+]
1.0 x 10^-12 = [H+]
pH = -log [H+]
pH = -log 1.0 x 10^-12
pH = 12.00

Question 11

what is the equivalence point?

Answer 11

the point in a titration at which the volume of one solution has reacted exactly with the volume of the second solution
[H+] = [OH-]
shown by the vertical region on a pH titration curve

Question 12

appearance of pH titration curve for strong acid and strong base

Answer 12

start at pH = 1
vertical region at pH = 3-11
end at pH = 13

Question 13

appearance of pH titration curve for strong acid and weak base

Answer 13

start at pH = 1
vertical region at pH = 3-8
end at pH = 11

Question 14

appearance of pH titration curve for weak acid and strong base

Answer 14

start at pH = 3
vertical region at pH = 5-10
end at pH = 13

Question 15

appearance of pH titration curve for weak acid and weak base

Answer 15

start at pH = 3
no vertical region
end at pH = 11

Question 16

how do we choose a suitable indicator?

Answer 16

the pH range/end-point of the indicator must coincide with the equivalence point of the titration

Question 17

what colour is phenolphthalein in acid/alkali?

Answer 17

acid = colourless
alkali = pink

Question 18

what colour is methyl orange in acid/alkali?

Answer 18

acid = red
alkali = yellow

Question 19

what is a buffer?

Answer 19

a mixture that minimises pH changes
on addition of small amounts of an acid or base
it is made up of a weak acid and its conjugate base

Question 20

how is a buffer solution prepared?

Answer 20

mix a weak acid and one of its salts together
or
react excess weak acid with a strong alkali

Question 21

what occurs when a small amount of acid (H+) is added to a buffer?
eg CH3COOH

Answer 21

CH3COOH (aq) → CH3COO- (aq) + H+ (aq)
conjugate base, CH3COO-, reacts with added H+ ions
equilibrium shifts left to minimise the effect of an increase in [H+]
most of the added H+ ions are removed

Question 22

when a small amount of base (OH-) is added to a buffer?
eg CH3COOH

Answer 22

CH3COOH (aq) → CH3COO- (aq) + H+ (aq)
H+ ions in the equilibrium react with the added OH-
equilibrium shifts right to minimise the effect of a decrease in [H+]
most of the H+ ions are restored

Question 23

equation for calculating [HA] : [A-]

Answer 23

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

Question 24

dissociation equation for carbonic acid in the blood

Answer 24

H2CO3 (aq) → H+ (aq) + HCO3 - (aq)