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