TOPIC 12 - ACID + BASE EQIUILIBRIA Flashcards
Bronsted-Lowry acid
eq
substance that is a proton donor (H+)
HA(aq) +H2O (l) –> H3O+ + A-
Bronsted-Lowry Base
eq
substance that can accept a proton - anything with a lone pair of electrons = dative covalent bond formed
B(aq) + H2O –> BH+ + OH-
strong acid vs weak acid
strong acid is able to completely dissociate.
weak acid partially dissociate (very slightly)
CH3COOH = CH3COO- + H+
weak bases
only slightly protonate in H20
NH3(aq) = NH4+ + OH-
examples of strong acids
HCl, H2SO4 (dibasic acid) , HNO3, HI, HBr, HClO, H3O+
examples of weak acids
HSO4- , H3PO4 (tribasic acid), HNO2, HF, HCN, NH4+, H2S, H2CO3
define pH
- log[H+] = concen of H+ in sol
- always 2d.p
find [H+]
10^-pH
amphoteric substances
H2O & NH3 can act as an acid and base
H2O + NH3 = NH4+ + H20
equil favour direction of where the stronger acid/base produces weaker acid and weaker base
monoprotic and diprotic acid dissociation
H2SO4 –> H+ + HSO4-
HSO4- –> H+ + SO4-
calc pH the slightly lower pH of first ionisation as not many H+ from HSO4- dissociate
for every change in 1 unit of pH the times of dilution/ concentration
x10^-1 (concen H+) = pH 1
x10^-2 = pH 2
x 10^1 (concen H+) = pH -1
but for weak acids the pH will increase less than 1 unit
acid + base reaction involves and eq
involves the transfer of a proton
acid + base -> conjugate acid (paired with base) + conjugate base
A + B –> BH+ + A-
to identify need to by looking both directions of the reaction
conjugate acid (BH+)
base (B) accepts a proton and forms a species
conjugate base (A-)
acid (HA) donates the proton and species is formed
Stronger the acid
the weaker the conjugate base
weaker the acid
the stronger the conjugate acid
if asked to find pH when giving the concen of a base
Ph + Poh = 14
rearrange after finding
Poh= -log[OH-]
then find Ph = ANS
e.g. 0.15 NaOH mol/dm3. find pH? [oh] = [0.15] Poh = -log[0.15] = 0.8239 = 14-0.8239 pH = 13.17
Ph + Poh = 14
only happens when
Kw = 1x10^-14
If Kw = 1x10^-15 then Ph + Poh = 15 etc
calc pH of weak acids
remember [H+]*mole ratio in equation
ALWAYS WRITE EQ
- partial dissociation HA = H+ + A- Ka (expresses how easily the acid releases H+) Ka = [H+] [A-]/[HA] due to 1:1 ratio Ka = [H+]^2/[HA]initial rearrange to find [H=] then pH can find pOH = 14 -pH
presentable/ manageable version of Ka
do not deal with standard form
pKa = -logKa Ka = 10^-pKa
the smaller the pKa
the greater the Ka greater the dissociation the stronger the acidity strength of acid increases pH<7
pH of weak acids (Ka) assumptions
- [H+] = [A-] at equilibrium they have dissociated at 1:1 ratio
- the amount of dissociation is small so assume inital concen of undissociated acid is CONSTANT and the acid that has been dissociated is CONSTANT = Ka
ionic product of water
Kc = [H+] [OH-] / [H20]
Kc*[H2O] = [H+][OH-]
constant as cocne on ions bigger than h2o =
Kw = [H+][OH-]
mol2/dm6
H20 dissociates slightly so endo so shit to right
25’C Kw=
1x10^-14 mol2/dm6
in pure water
[H+] = [OH-] are equal = 1x10^-14
[H+] = 1x10^-7 = [OH-]
or [H+]^2 = Kw
= pH of water = 7
pKw
-logKw
calc pH of strong base
either Kw or Poh+Ph = 14
depending on Kw
find pH from concen of diprotic and triprotic acid
need to multiply concen of acid by the no. of protons to find [H+]
Why is pure water still neutral even if pH does not equal 7?
Kw affected by temp similar to equil
does not mean pH = 7
but same
[H+] = [OH-]
enthalpy change of neutralisation for strong acids and strong base
it will be similar = -57kJ/mol because the reaction btw them they are almost fully dissociated in aq sol
enthalpy change of neutralisation for weak acids or weak base
is less than -57.6kJ/mol and there is a major difference as majority of acid molecules are undissociated so require less energy to dissociate them. So less heat energy given when CH3COOH & HCN are neutralised by NaOH
HF is a weak acid but has a enthalpy change of neutralisation higher than -57kJ/mol why?
so more exothermic as the enthalpy of hydration releases more heat energy than the dissoiation molecules takes in
pH of sol of strong acid and strong base?
n(H+) = cxvx no.moles of H N(OH-) = cxv which one is in excess xs (OH-) = n(OH-) - n(H+) [OH-] = xs OH- / V Poh = -log[OH-] pH calc
What ion causes a solution to be acidic (2 marks)
H+ ions release in water combining with H2O to form H3O+ oxonium ions.
What physical factors affect the value of Kw and how do they affect it
Temperature only - if temperature is increased, equilibrium moves to the right so Kw increases and the ph of the pure water decreases
What is the relationship between pH and concentration of H+?
Lower pH = higher concentration of H+
What is the difference between concentrated and strong
Concentrated means many mol per dm3, strong refers to amount of dissociation
what is a buffer solution?
Buffer solution is a system that minimizes pH changes when addition of small amounts of an acid or base
how buffers resist changes to pH when small amount of acid added?
- write equil
- influx of H+
- pushes equil to left by reacting with reservoir of propanoate ions - remove H+
how buffers resist changes to pH when small amount of base added?
- added OH-
- reacts with acid to form -oate ions (A-) and water
- or reacts with H+ to form water
- H+ concen drops - equil shifts right
- dissociating more acid to upkeep he H+ concen therefore no change in pH
composition of a buffer
make a sol of a weak acid + salt of the weak acid
- achieved by neutralise half of weak acid with an alkali - forms weak acid salt
opposite forbasic buffer sol = weak base + soluble salt
facts about buffers
- 2 equations
- weak acid partially dissociates
- salt of the acid completely dissociates
- in Ka the total [A-] is the total [] present in the system from both equations
2 assumptions abt buffers
- considering [A-] is very small almost negligible due to partial dissociation. so total [A-] comes from salt of the weak acid
- NEVER Ka = [H+]^2 /[HA]
Buffer action in blood
pH = 7.4
H2CO3- carbonic acid (H+ + HCO3-) = CO2 + H20
addition of acid/base change equil
carbonic acid = buffer
What is titration?
Addition of acid/base of known titration to a base/acid of unknown titration to determine the concen. An indicator is used to show that neutralization has occurred
- clamp stand, burette in that is acid/alkali of known concen, conical flask, whit tile
- calibrate and improve by maintaining constant temp
calculate Ka for a weak acid from experimental data given the pH of a
solution containing a known mass of acid
m (acid) = 0.49g pH = 3 equil Mr(acid) = 122g/mol n(acid) [acid] [H+] = 10^-3.00 = [A-] calc Ka = [H+]^2/[HA] compare with data book value
obtain a pH titration curve
- add smaller volumes and more accurate curve produced
equivalence point
the point at which the exact vol of base has been added to just neutralise the acid. vertical point on curve and usually a large and rapid change in pH except in weak-weak titration
end point
he vol of acid/lkali added when the indicator just changes colour.
when does the equivalence point = end point
if the right indicator used
what are the properties of a good indicator for a reaction? 3 marks
- sharp colour change and not gradual - only 1 drop can cause change in colour
- end point must be the same as the equivalence point or titration is wrong answer
- distinct colour change so obvious when end point reached
half neutralisation point
when half of volume added at equivalence point
- at half equivalence point pH = pKa = buffer
titration curve and how to draw
y-axis = pH x-axis = vol of _ added
- strong acid + strong base
- strong acid + weak base
- weak acid + strong base
- weak acid + weak base
indicator look at data book - calc vol of half equivalence point
