3.9 acid-base equilibria Flashcards
an acid is a proton (donor/acceptor)?
donor
(H+ donor)
a base is a proton (donor/acceptor)?
acceptor
(H+ acceptor)
it is possible that a species that behaves like an acid in one reaction will behave like a base in another
in acid-base reactions we get conjugate acid-base pairs
e.g NH3 + H2O —> NH4+ + OH-
base acid CA CB
- nitric acid, HNO3, is a strong Bronsted-Lowry acid
- nitrous acid, HNO2, is a weak Bronsted-Lowry acid
- A student suggests that an acid-base equilibrium is set up when nitric acid is mixed with nitrous acid
- complete the equation for the equilibrium that would be set up and label the conjugate acid-base pairs for:
HNO3 + HNO2 ⇌ ____ + _____
HNO3 + HNO2 ⇌ NO3 - + H2NO2 +
HNO3 = acid
HNO2 = base
NO3 - = conjugate base
H2NO2 + = conjugate acid
what are strong acids?
- acids that fully dissociate
- HA —> H+ + A-
what are weak acids?
- acids that only partially dissociate (in a reversible reaction)
- HA ⇌ H+ + A-
what does the degree of dissociation of a weak acid depend on?
its acid dissociation constant, Ka
express Ka for HA ⇌ H+ + A-
Ka = [H+][A-] / [HA]
for strong acids, HA —> H+ + A- and the concentration of HA = H+
for weak acids, HA ⇌ H+ + A-, with conc of HA 0.1M, how do you find out the concentration of H+?
Ka = [H+][A-] / [HA]
- [H+] = [A-]
- [HA] - approximately 0.1M because some has split already
so: (for weak acids)
Ka = [H+]^2 / [HA]
what are some examples of weak acids?
- ethanoic acid
- any carboxylic acid
need to remember the pH calculations for AS
what are the 2 assumptions we make to say that for weak acids, Ka = [H+]^2 / [HA]?
- assume [H+] = [A-] because they have dissociation according to a 1:1 ratio
- assume the initial concentration of the undissociated acid remains constant because the amount of dissociation of a weak acid is so small
- ethanoic acid CH3COOH
- ethanoic acid is a weak acid with an acid dissociation constant, Ka of 1.75×10^-5 moldm^-3 at 25°C
- the student uses a pH meter to measure the pH of a solution of CH3COOH at 25°C
- the measured pH is 2.440
- calculate the concentration of ethanoic acid in the solution [3]
Ka = 1.75x10^-5
Ka = [H+] / [HA]
[H+] = 10 ^ -2.44 = 3.63 x10^-3
1.75x10^-5 = (3.63)^2 / [HA]
[HA] = 0.753
the dissociation of water equation:
H2O ⇌ H+ + OH-
- water is always slightly dissociated into hydrogen ions and hydroxide ions
the concentration of water [H2O] is constant, such that it can be incorporated into a new constant Kw, where at 298K:
Kw=
Kw = [H+][OH-]
= 1.0 x10^-14 mol^2dm^-6
- this is the ionic product of water
- this means that if the conc of H+ ions increases (e.g by adding acid to the solution), the conc of OH- ions must decrease proportionally
any solution in which the concentrations of H+ and OH- are equal are said to be what?
neutral
[H+] = [OH-]
the dissociation of water is an (endothermic/exothermic) reaction?
endothermic
1cm^3 = 1g H2O
Mr= 18.02
mass = 1g
n = 0.055
v = 0.001dm^3
so conc = 55.5
[H2O] = 55.5 moldm^-3
- constant in any aq solution
what is the value of the Kw of water?
1 x10^-14 mol^2dm^-6
- tiny so equilibrium very to RHS
- ∴ H2O hardly dissociated
neutrality rests on the ____ of the water
temperature
Kw = [H+][OH-]
- if non-neutral i.e basic, Kw = ?
- if neutral bc [H+]=[OH-], Kw = ?
- Kw = [H+][OH-]
- Kw = [H+]^2 (pure)
what is a strong base?
one in which fully dissociated into its ions in water
(strong bases in water are rare as water is a very weak acid and will not readily give up protons to other species)