acids + bases Flashcards
(41 cards)
define bronsted-lowry acid
proton donor
define bronsted-lowry base
proton acceptor
define Lewis acid
e- pair acceptor
define lewis base
e- pair donor
when acid donates proton species formed is
conjugate base of acid
why is a log scale used for ph
conc of H+ in aq soln covers v wide range
at dif temps to 25 deg C pH of pure water changes, how to predict this change?
- use le chateliers principle
- dissociation of water = endothermic so increase temp - favours forward - pushes equilibrium to right - greater conc of H+ - lower pH
what does a larger Ka mean
strong acid
how to calc pKa
-log Ka
what assumptions need to be made when calculating pH of weak acid
1) [H+ (aq)]eqm = [A- (aq)] eqm because they have dissociated according to a 1:1 ratio.
2) As the amount of dissociation is small, assume that the initial concentration of the undissociated acid has remained constant.
So [HA (aq) ] eqm = [HA(aq) ] initial
15cm3 of 0.5 mol dm-3 HCl is reacted with 35cm3 of 0.55 mol dm-3 NaOH. Calculate the pH of the resulting mixture.
- MOLES OF ACID: Moles HCl = mol H+ = conc x vol = 0.5 x 0.015 = 0.0075mol
- MOLES OF BASE: Moles NaOH =mol OH- = conc x vol = 0.55 x 0.035 = 0.01925
- WHICH IS IN EXCESS: Moles of OH- in excess = 0.01925 – 0.0075= 0.01175
IF ALKALI EXCESS - WORK OUT NEW CONC OF XS OH- IONS: [OH-] = moles excess OH-/ total volume (dm3)
= 0.01175/ 0.05 = 0.235 mol dm-3 - USE Kw TO FIND OUT H+ CONC
[H+] = Kw /[OH– ]
= 1x10-14 / 0.235 = 4.25x10-14 - FIND pH
pH =–log[H+]
= -log 4.25x10-14
= 13.37
35cm3 of 0.5 mol dm-3 H2SO4 is reacted with 30cm3 of 0.55 mol dm-3 NaOH. Calculate the pH of the resulting mixture.
- CALC MOLES ACID: Moles H2SO4 = conc x vol = 0.5 x 0.035 = 0.0175mol
- NEED TO x2 AS DIPROTIC ACID: Moles H+ = 0.0175 x2 = 0.035
- CALC MOLES BASE: Moles NaOH = mol OH- = conc x vol = 0.55 x 0.030 = 0.0165
- CALC EXCESS OF MOLES: Moles of H+ in excess = 0.035 -0.0165 = 0.0185
- CALC CONC OF H+:
[H+] = moles excess H+ total volume (dm3)
= 0.0185/ 0.065 = 0.28 mol dm-3 - CALC pH
pH = –log[H+] = -log 0.28
= 0.55
55cm3 of 0.50 mol dm-3 CH3CO2H is reacted with 25cm3 of 0.35 mol dm-3 NaOH. Calculate the pH of the
resulting mixture. Ka is 1.7x10-5moldm-3
- CALC MOLES OF OG ACID: Moles CH CO H = conc x vol =0.5x 0.055 = 0.0275mol 3 2
- CALC MOLES OF BASE ADDED:
Moles NaOH = conc x vol = 0.35 x 0.025 = 0.00875 - WHICH IN XS?
Moles of CH3CO2H in excess = 0.0275-0.00875 = 0.01875 (as 1:1 ratio) - IF ACID IN XS: WORK OUT NEW CONC OF XS HA
[CH3CO2H ] = moles excess CH3CO2H total volume (dm3)
= 0.01875/ 0.08 = 0.234M - WORK OUT CONC OF SALT FORMED (A-):
[CH3CO2- ] = moles OH- added/ total volume (dm3)
= 0.00875/ 0.08 = 0.109M
REARRANGE: Ka =[H+][CH3CO2-] [H+]=K1 x[CH CO H]/[CH CO -] [ CH3CO2H ]
=1.7x10-5 x0.234/0.109 = 3.64 x 10-5
FIND pH
pH = – log [H+] =-log3.64x10-5
= 4.44
what to do when work out pH of WA at half equivalence
When a weak acid has been reacted with exactly half the neutralisation volume of alkali, the above calculation can be simplified
Ka = [H+] [CH3CO2- ]/ [ CH3CO2H ]
At half neutralisation we can make the assumption that [HA] = [A-]
So [H+] = Ka
And pH = pKa
how to work out new conc of diluted acid/alkali
[H+] = [H+]old x old volume/ new volume
[OH–] = [OH–]old x old volume/
new volume
what does a buffer soln do
where the pH does not change significantly if small amounts of acid or alkali are added to it.
how is an acidic buffer soln made
from a weak acid and a salt of that weak acid (made from reacting the weak acid with a strong base)
Example : ethanoic acid and sodium ethanoate
CH3CO2H (aq) and CH3CO2- Na+
how is a basic buffer soln made
from a weak base and a salt of that weak base (made from reacting the weak base with a strong acid).
Example :ammonia and ammonium chloride NH3 and NH4+Cl-
how do buffer soln work
In an ethanoic acid buffer
CH3CO2H (aq) <-> CH3CO2- (aq) + H+ (aq)
in buffer soln what is in higher conc compared to pure acid
the salt ion
what happens when small amounts of acid r added to Ethan oil acid buffer?
Then the above equilibrium will shift to the left removing nearly all the H+ ions added,
CH3CO2- (aq) + H+ (aq) CH3CO2H (aq)
As there is a large concentration of the salt ion in the buffer, the ratio [CH3CO2H]/ [CH3CO2-] stays almost constant, so the pH stays fairly constant.
what happens when small amounts of alkali are added to ethanoic acid buffer
The OH- ions will react with H+ ions to form water.
H+ + OH - H2O
The equilibrium will then shift to the right to produce more H+ ions.
CH3CO2H (aq) CH3CO2- (aq) + H+ (aq)
Some ethanoic acid molecules are changed to ethanoate ions but as there is a large concentration of the salt ion in the buffer, the ratio [CH3CO2H]/ [CH3CO2-] stays almost constant, so the pH stays fairly constant.
Calculate the pH of a buffer made from 45 cm3 of 0.10 mol dm-3 ethanoic acid and 50cm3 of 0.15 mol dm-3 sodium ethanoate (Ka = 1.7 x 10-5)
Calculate the moles of both solutions
Moles ethanoic = conc x vol = 0.1 x 0.045 = 0.0045 mol
Moles sodium ethanoate = conc x vol = 0.15 x 0.050 = 0.0075 mol
[H+(aq)] = Ka (aq) x
[HA ] /[A- (aq) ]
^ We can enter moles of acid and salt straight into the equation as they both have the same new final volume
[H+(aq)] = 1.7 x 10-5 x 0.0045
[H+(aq)] = 1.02x 10-5 0.0075
pH = – log [H+]
= -log 1.02x 10-5
= 4.99
A buffer solution is made by adding 1.1 g of sodium ethanoate into 100 cm3
of 0.40 mol dm-3 ethanoic acid. Calculate its pH. (Ka =1.7 x10-5 mol dm-3 )
Calculate the moles of both solutions
Moles ethanoic = conc x vol = 0.4 x 0.1 = 0.04 mol
Moles sodium ethanoate = mass/Mr = 1.1/82 [H+(aq)] = 0.0134 mol
[H+(aq)] = Ka (aq) x [HA ]/ [A- (aq) ]
[H+(aq)] = 1.7 x 10-5 x 0.04/0.0134
[H+(aq)] = 5.07x 10-5
pH = – log [H+]
= -log 5.07x 10-5
= 4.29
