5.1.3 Acids, Bases and Buffers Flashcards
what is a Bronsted-Lowry acid
a proton (H+) donor
what is a Bronsted-Lowry base
a proton (H+) acceptor
what are conjugate base pairs
consist of 2 species that can be interconverted by transfer of a proton
in the dissociation of HCl, what forms conjugate base and what forms conjugate acid
HCl → H+ + Cl-
1) in forward direction, HCl releases a proton and forms conjugate base Cl-
2) in backwards reaction, Cl- accepts a proton and forms HCl
what reaction is always taking place in a neutralisation
H+ + OH- → H2O
what would be the acid base equilibrium of the dissociation of HCl
HCl + OH- ⇌ H2O + Cl-
what are the conjugate acid pairs in the acid base equilibrium of dissociation of HCl
HCl + OH- ⇌ H2O + Cl-
FORWARD DIRECTION:
- HCl donates H+, so acid 1
BACKWARD DIRECTION
- Cl- accepts H+, so base 1
FORWARD DIRECTION+
- OH- accepts H+, so base 2
BACKWARD DIRECTION:
- H2O donates H+, so acid 2
what is a hydronium ion
H3O+(aq)
- ACTIVE INGREDIENT IN ANY AQUEOUS ACID
where is the hydronium ion made
- in aqueous solutions of dissociation reactions, where a proton needs to be transferred from acid to base, you MUST have water present
- so a proton is transferred to water, forming H3O+
rewrite the HCl acid-base equilibrium, but this time using the H3O+ ion
HCl + H2O ⇌ H3O+ + Cl-
how is the equation for neutralisation simplified from using the H3O+ ion
H3O+ + OH- → 2H2O
H+ + OH- → H2O
what does the ____basic bit of an acid tell you
the total number of H+ ions in the acid that can be replaced per molecule in an acid-base reaction
- e.g. monobasic, dibasic, tribasic
why is the ____basic bit of an acid useful
- used to work out neutralisation equations
- each of the H+ must be replaced, so helps you find out how much base you need
- also able to find out volumes too
what is the role of H+ ions in acids
- the active species of acids that is reacting
- can be emphasised through writing ionic equations, and you’ll find for any acid with the same basic, gives you the same ionic equation
what is the ionic equation for 2HCl + Mg
2H+ + Mg → Mg2+ + H2
what are spectator ions
ions that do not change in a reaction, always cancelled out
what is a redox reaction that acids go through
acid + base → salt + hydrogen
what is the ionic equation basic layout for acid + carbonate
2H+ + CO32- → salt + H2O + CO2
what is the ionic equation basic layout for acid + base
2H+ + MgO → Mg2+ + H2O
what is the ionic equation basic layout for acid + alkali
H+ + OH- → H2O
- as alkali is soluble in aqueous solutions, so its ions will cancel out too
what is pH
a measure of hydrogen ion H+ concentration
how is H+ ion concentration converted to pH
10⁻¹⁴ goes to pH 14
- so a low [H+(aq)] is a high pH
- and a high [H+(aq)] is a low pH
what pHs indicate what
pH>7 : alkali
pH<7 : acid
pH=7 : neutral
how can pH be measured
using a pH meter or indicators, either paper or universal
what type of scale is pH
logarithmic
- one pH number change is a x10 jump in concentration of H+ ions
- e.g. pH 1 has 10x more H+ ions than pH 2, and so on
how can you convert between pH and [H+]
pH = -log[H+]
[H+] = 10⁻ᵖᴴ
what do strong, monobasic acids do in aqueous solutions, and what does this mean
COMPLETELY dissociate
- HA → H+ + A-
- one mole of HA turns into one mole of H+
- so [HA] = [H+]
what is the equation for finding [H+] in strong acids
[H+]=[HA]
- concentration of acid is equal to concentration of H+ ions
how can you figure out pH changes based on dilution
- through using [H+]=[HA] formula for both
- we know pH 1 has x10 [H+] than pH2, but this is not a set scale
what is the difference between strong and weak acids
strong acids: completely dissociate in aqueous solutions (HCl)
weak acids: partially dissociate in aqueous solutions (CH3COOH) ⇌
what constants can be used for acid-base equilibria
- special equilibrium constants
- but all just versions of Kc
- where [products]/[reactants]
what is the dissociation of weak acids formula
HA ⇌ H+ + A-
what is the acid dissociation constant called
Ka
- used to find out [H+] of weak acids
what is the Ka
[H+][A-]/[HA]
what are the units of Ka
moldm-3
when will Ka change
- ONLY with temperature, as will all Kc
- set at standard 25 degrees celsius
what does a larger KA value show
- the equilibrium is further to the right, so greater [H+]
- greater dissociation
- STRONGER the acid
why is Ka often difficult to deal with
- often get numbers with negative indices
- difficult to compare, so convert to the negative log instead of pKa
- is more manageable than Ka
- and easier to compare relative strengths
what is the conversion of Ka to pH
pKa = logKa
Ka = 10⁻ᵖᴷᵃ
what is the relationship between strength of acid, Ka and pKa
- strong acid
- large Ka
- small pKa
at equilibrium, how is [HA] and [H+] related for strong and weak acids
strong acid : [H+]=[HA]
weak acid : [H+]≠[HA] , but rather HA ⇌ H+ + A-
what does the concentration of H+ formed at equilibrium depend on
- the concentration of the acid
- the Ka acid dissociation constant
rewrite Ka using equilibrium values
[H+]equi[A-]equi/[HA]equi
- where [HA] equi can be rewritten as [HA]start-[H+]equi
- where when KA dissociates, H+ and A- are mad in equal quantities
what are the two approximations you make when identifying the value of Ka of an acid
1) H+ and A- are equal, so [H+]=[A-]
(although are the same, some H+ is formed via the dissociation of water, but will be extremely small and can be neglected
2) assume [HA]»[H+], as dissociation of weak acid is small, so can neglect any decrease in concentration of HA from dissociation, so [HA]start=[HA]equi
what is the Ka formula produced once approximations have been accounted for
[H+]²/[HA]
how can you work backwards from Ka to figure out [H+]
[H+] = √Kax[HA]
- from this, can calculate pH via -log[H+]
how can you determine Ka experimentally
- prepare a standard solution of a weak acid of known concentration
- measure the pH using a meter
- work out [H+] using 10⁻ᵖᴴ
what are the limits of the approximations we make when using Ka formula
1) assume dissociation of water is negligible, so [H+]=[A-]
- at 25 degrees, dissociation of water = 10⁻⁷, so if acid has pH above 6, it becomes significant
- so approximation breaks down for: VERY WEAK OR DILUTE ACIDS
2) assume [HA]»[H+], so only works for weak acids with a small Ka value, as otherwise [HA] would be drastically different from [HA] at equilibrium ([HA]-[H+])
- so approximation breaks down for stronger weak acids with Ka>10⁻² or very dilute solutions
what happens during the ionisation of water
- ionises slightly, and acts as both an acid and a base
- H2O + H2O ⇌ H3O+ + OH-
- simplifies to H2O ⇌ H+ + OH-
what would Ka be if we treat water as a weak acid
[H+][OH-]/[H2O]
whats special about the dissociation of water
- it is VERY SMALL, so [H2O] can be treated as a constant
what does Ka of water rearrange to
Ka x [H2O] = [H+][OH-]
Ka x [H2O] is Kw
what is Kw
the ionic product of water
what is the equation of Kw
Kw = [H+][OH-]
when does Kw vary
varies with temperature, but at 25 degrees celsius, is equal to 1x10⁻¹⁴
how can the pH of pure water be identified at 25 degrees celsius
- on dissociation, is neutral, so produces same number of [OH-] as [H+]
- so [OH-]=[H+]
- Kw=[H+]²
- = 1 x 10⁻¹⁴
- rearrange for H+ and put into pH formula and you get answer of 7
what does Kw control
equilibrium constant that controls the concentrations of H+ and OH- in aqueous solutions
- if acid: [H+]>[OH-]
- if alkali: [H+]<[OH-]
- if neutral: [H+]=[OH-]
- Kw has a value of 1 x 10⁻¹⁴, which is able to control these relative concentrations
for whole numbered pHs, how can Kw be used quickly
the indices just need to add up to -14, so if you know one, you know the other
what is an alkali
a soluble base that releases OH- ions in aqueous solutions
what is a strong base
COMPLETELY DISSOCIATES
- AOH → OH- + H+
- AOH and OH are both one mole
- so [AOH]=[OH-]
- where A is any metal
how can you find pH of a strong base
1) from concentration of acid, you know concentration of OH- as they are equal
2) can sub this into Kw along with Kw’s value
3) find out concentration of H+ from this
4) sub into pH=-log[H+]
what is a weak acid
equilibrium is positioned well to the left, only partially dissociate and release OH- ions in aqueous solutions
