physical - acids&bases Flashcards
defining an acid
a Brønsted-Lowry acid
a proton donor
a Brønsted-Lowry base
a proton acceptor
water as a base
HCl + H2O
HCl(g) + H2O(l) –> H3O+(aq)+ Cl-(aq)
HCl donates a proton to H2O - acid
H2O accepts a proton from HCl - base
neutralisation
occurs when there is a transfer of a proton involved
what is H3O+
oxonium ion
water as an acid
NH3 + H2O
NH3(g) + H2O(l) –> OH-(aq) + NH4+(aq)
H2O donates a proton to NH3 - acid
NH3 accepts a proton from H2O - base
the ionization of water
H2O(l) ⇌ H+(aq) + OH-(aq)
or
H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq)
what is Kc
expression of the equilibrium of ionisation of water
Kc= [H+ (aq)][OH- (aq)] / [H2O(l)]
rearrangement of Kc
- because [H2O (l)] is much bigger than the concentrations of the ions, we assume its value is constant and make a new constant Kw
Kc x [H2O (l)] = [H+ (aq) ][OH- (aq)]
–> Kw = [H+ (aq) ][OH- (aq) ]
what is Kw
- the ionic product of water
Kw = [H+ (aq) ][OH- (aq) ] - and at 298K, the value of Kw for all aqueous solutions is 1x10^-14 mol2dm-6
= Kw = [1x10^-7 mol dm^-3][1x10^-7 mol dm^-3] - H2O dissociates to give one H+ and one OH-
factor that affects Kw
- temperature only
- if temperature increase, equilibrium moves to the right so Kw increases and pH of pure water decrease
why is pure water still neutral even when the pH does not = 7
[H+] = [OH-]
strong bases
- a strong base dissociates completely in water, releasing OH- ions
how to calculate pH of strong bases
- Kw = [H+ (aq) ][OH- (aq) ]
- [H+] = Kw / [OH-]
- we know Kw = 1.00 x 10^-14 in constant conditions
- [OH-] will vary with every questions
solve: calculate the pH of a 0.006 mol dm^-3 solution of sodium hydroxide
- kw = [H+][OH-] = 1 x 10-14mol-2 dm-3
- [H+] = Kw / [OH-]
- [H+] = 1x10^-14 / [0.006]
- [H+] = 1.67x10^-12
- pH = -log[H+]
- pH = -log[1.67x10^-12]
- pH = 11.78
