Acid-base pH III Flashcards
Polyfunctional acids & bases
- Dibasic acid
-2 ionisable groups
Strong acid
Tribasic
3 ionisable group
Dicarboxylic acids
extra COOH: e- withdrawing stabilises conjugate base
Diamines
-NH2 and -NH3- e- withdrawing
Extra -NH2
not very marked effect on 1st pKa1 reduction
Difference between pKa1 and pKa2 _______ as -CH2- seperating COOH ________
diminishes
increases
After removing the first H from COOH; pKa2 < pKa1 COO- electron-donating ______ the dianion.
destabilise
Extra COOH: ______ acidity
increase
lower the pKa ______ the acidity
higher
2nd pKa is always bigger. Why?
1st ionisation COOH increase acidity of compound but once it has been ionised to COO- it will decrease acidity of compound.
Extra -NH3+
reducing a lot of the pKa2
Examples of oxoacids;
HNO3, H2SO4, H3PO4
they contain OH groups that ionise (produce H+) in water and (n-m) doubly boded O
Na2O is a _______ oxide
basic = reacts with water to make NaOH solution
SO2 is an ______ oxide
reacts with water to make H2SO3 (eq w/ HSO3-)
Basic oxides reacts with ____
Acidic oxides reacts with ____
acids
bases
titration curve dibasic acid-strong base = 2 moles base and 1 mol acid: DRAW CURVE
1st equivalence point at pH4 and 2nd at pH 9
2 independent reactions with base as its dibasic (2pKa)
Equivalence point; acid-base –>
equal stoichiometric amount
reaction completed when…
all acid has been converted to its conjugate base
Speciation curve
look at lecture slides/google
how to use pKa values
predicts outcome of acid-base interactions
acid-base interactions
the formation of weaker acid and weaker base (most stable being favoured at equilibrium)
if there is a large pKa difference =
reaction irreversible
because strong acid and base is being used.
conjugate base benzoate is a water-soluble anion = push reaction to the right.
Reason for preparing a salt for a non water soluble acidic drug.
methyl-amine + water —> ammonium chloride and water
methyl-amine will become protonated (added H+)
pKa increases from 1.74 (water) to 10.7 (NH4Cl) because it is a weaker acid
why does equilibrium favour the right side?
there is a large difference in pKa
NaNH2: formation of NH3
pKa 5.7 stronger acid (H2O) give a H+ tpo make a weaker base OH- and a weaker acid with a pKa 38
H2O donate H+ to any _____ stronger than OH-
base
to use bases that are stronger than OH- :
solvent weaker acid than H2O
Examples of weak acids?
Hydrocarbons (pKa 50)
Ammonia (pKa 38)
Amino acids are _____ compounds;
amphoteric
amphoteric compounds?
act as proton donor and as a proton acceptor (acid/base depending on conditions)
ALL amphiprotic substances are _________ - but reverse isn’t true
amphoteric
Zwitterion;
doubly charged form
COOH protonates NH2 at pH 7, amino acids having neutral R will exist predominantly as
zwitterion (internal salt)
ammonium cation > zwitterion net charge is ____> aminocarboxylate anion
zero
amino acids are _____ compounds - therefore they have no pH existence
ionic
Isoelectric point pI; pH at which concentration of the zwitterion is max pH which the concentrations of cationic and anionic forms are _____
equal
pI=pH=
pKa1 + pKa2 / 2
equation for Ka1?
ka1 = [H+] [zwitterion] / [cation]
equation for Ka2?
Ka2 = [H+] [anion] / [zwitterion]
At pI: [cation] = [anion]
[cation] = [H+] [zwitterion] / Ka1 = [anion] = Ka2 [zwitterion] / [H+]
the isoelectric point (pI)
the pH at which a particular molecule carries no net electrical charge
Ka1 x Ka2 = [H+]2 so what is the equation for pH
pH = pK1 + pKa2 / 2
pI: pKa2 (cation) = pKa3 (dipolar ion) / 2
examples…
amino acids; positively charged at low (acidic) pH, negatively charged at high (basic) pH and zwitterionic at neutral pH
implications fro oral absorption and bioavailability of amino acids from diet
bioavailability nutrition?
the efficiency with which a dietary component is used. systematically through normal metabolic pathways
Ionised species (substances with a charge): veryy low lipid solubility
unable to permeate through membranes
only non-ionised drug is usally able to cross membranes
When using H-H equation =
pH = pKa + log [base] / [acid]
if we want pH = pKa for drug (weak acid)
log [base] / [acid]= log [ionised] / [non-ionised] = 0
if shift in pH by one unit other side of pKa = must change the ratio of ionised to non-ionised forms by factor of 10
pH = pKa + log [ionised] / [non-ionised]
pH = pKa + 1 [ionised / [non-ionised] = 10 10 / 1
which means [ionised] 10 times
pH = pKa + log [ ionised] / [non-ionised]
pH = pKa - 1 [ionised] / [ non=ionised] = 0.1
1/10 [non-ionised] 10 times
calculate fraction of the total dose that is ionised of any pH if the pKa is known
fraction [ionised] = 1 / 1 + antilog (pKa - pH)
%ionised = 100 / 1+ antilog (pKa - pH)
weak bases
fraction [ionised] = 1 / 1 + antilog (pH - pKa)
% ionised = 100 / 1 + antilog (ph - pKa)
