Acids and Bases Flashcards
Bronsted-Lowry acid
donate a proton
Bronsted-Lowry base
accept a proton
acid base reaction
HCl (g) + H2O (l) -> H3O+ (aq) + Cl-
ph
pH = - log [H+]
strong acid
completely dissociate
[H+]
10-pH
Ionic product for water
Kw = [H+ (aq) ][OH- (aq) ]
assume concentration of water is constant as there is so much of it compared to the ions
dissociation of water- prediction
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
pH of a strong base
use Kw expression
pH of a weak acid assumptions
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.
pH of a weak acid
Ka = H2 / HA
neutralisation- excess acid
work out H+
pH = –log[H+]
neutralisation- excess base
work out OH-
[H+] = Kw /[OH– ]
pH = –log[H+]
pH at half equivalence
At half neutralisation we can make the assumption that [HA] = [A-]
so [H+] = Ka
pH = pKa
Ka from pKa
Ka = 10-pKa
buffer solution
pH does not change significantly if small amounts of acid or alkali are added to it
acidic buffer
made from a weak acid and a salt of that weak acid (made from reacting the weak acid with a strong base).
basic buffer
made from a weak base and a salt of that weak base (made from reacting the weak base with a strong acid).
small amounts of acid is added to the buffer
then the above equilibrium will shift to the left removing nearly all the H+ ions added
small amounts of alkali is added
OH- ions will react with H+ ions to form water.
The equilibrium will then shift to the right to produce more H+ ions.
pH of a buffer
assume the [A-] concentration is due to the added salt only
H+ = Ka (HA/A-)
