Acids and Bases

Question 1

Bronsted-Lowry acid

Answer 1

donate a proton

Question 2

Bronsted-Lowry base

Answer 2

accept a proton

Question 3

acid base reaction

Answer 3

HCl (g) + H2O (l) -> H3O+ (aq) + Cl-

Question 4

ph

Answer 4

pH = - log [H+]

Question 5

strong acid

Answer 5

completely dissociate

Question 6

[H+]

Answer 6

10-pH

Question 7

Ionic product for water

Answer 7

Kw = [H+ (aq) ][OH- (aq) ]

assume concentration of water is constant as there is so much of it compared to the ions

Question 8

dissociation of water- prediction

Answer 8

The dissociation of water is endothermic so increasing the temperature would push the equilibrium to the right giving a bigger concentration of H+ ions and a lower pH.

increase temp = lower pH

Question 9

pH of a strong base

Answer 9

use Kw expression

Question 10

pH of a weak acid assumptions

Answer 10

assume :
H+ = A- because they have dissociated according to a 1:1 ratio.
[HA (aq) ] eqm = [HA(aq) ] initial
amount of dissociation is small, assume that the initial concentration of the undissociated acid has remained constant.

Question 11

pH of a weak acid

Answer 11

Ka = H2 / HA

Question 12

neutralisation- excess acid

Answer 12

work out H+

pH = –log[H+]

Question 13

neutralisation- excess base

Answer 13

work out OH-

[H+] = Kw /[OH– ]

pH = –log[H+]

Question 14

pH at half equivalence

Answer 14

At half neutralisation we can make the assumption that [HA] = [A-]
so [H+] = Ka
pH = pKa

Question 15

Ka from pKa

Answer 15

Ka = 10-pKa

Question 16

buffer solution

Answer 16

pH does not change significantly if small amounts of acid or alkali are added to it

Question 17

acidic buffer

Answer 17

made from a weak acid and a salt of that weak acid (made from reacting the weak acid with a strong base).

Question 18

basic buffer

Answer 18

made from a weak base and a salt of that weak base (made from reacting the weak base with a strong acid).

Question 19

small amounts of acid is added to the buffer

Answer 19

then the above equilibrium will shift to the left removing nearly all the H+ ions added

Question 20

small amounts of alkali is added

Answer 20

OH- ions will react with H+ ions to form water.

The equilibrium will then shift to the right to produce more H+ ions.

Question 21

pH of a buffer

Answer 21

assume the [A-] concentration is due to the added salt only

H+ = Ka (HA/A-)