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PMP-102 > Acid-base pH III > Flashcards

Acid-base pH III Flashcards

1
Q

Polyfunctional acids & bases

A
  • Dibasic acid

-2 ionisable groups

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2
Q

Strong acid

A

Tribasic

3 ionisable group

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3
Q

Dicarboxylic acids

A

extra COOH: e- withdrawing stabilises conjugate base

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4
Q

Diamines

A

-NH2 and -NH3- e- withdrawing

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5
Q

Extra -NH2

A

not very marked effect on 1st pKa1 reduction

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6
Q

Difference between pKa1 and pKa2 _______ as -CH2- seperating COOH ________

A

diminishes
increases

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7
Q

After removing the first H from COOH; pKa2 < pKa1 COO- electron-donating ______ the dianion.

A

destabilise

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8
Q

Extra COOH: ______ acidity

A

increase

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9
Q

lower the pKa ______ the acidity

A

higher

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10
Q

2nd pKa is always bigger. Why?

A

1st ionisation COOH increase acidity of compound but once it has been ionised to COO- it will decrease acidity of compound.

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11
Q

Extra -NH3+

A

reducing a lot of the pKa2

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12
Q

Examples of oxoacids;

A

HNO3, H2SO4, H3PO4

they contain OH groups that ionise (produce H+) in water and (n-m) doubly boded O

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13
Q

Na2O is a _______ oxide

A

basic = reacts with water to make NaOH solution

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14
Q

SO2 is an ______ oxide

A

reacts with water to make H2SO3 (eq w/ HSO3-)

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15
Q

Basic oxides reacts with ____
Acidic oxides reacts with ____

A

acids
bases

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16
Q

titration curve dibasic acid-strong base = 2 moles base and 1 mol acid: DRAW CURVE

A

1st equivalence point at pH4 and 2nd at pH 9
2 independent reactions with base as its dibasic (2pKa)

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17
Q

Equivalence point; acid-base –>

A

equal stoichiometric amount

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18
Q

reaction completed when…

A

all acid has been converted to its conjugate base

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19
Q

Speciation curve

A

look at lecture slides/google

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20
Q

how to use pKa values

A

predicts outcome of acid-base interactions

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21
Q

acid-base interactions

A

the formation of weaker acid and weaker base (most stable being favoured at equilibrium)

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22
Q

if there is a large pKa difference =

A

reaction irreversible

because strong acid and base is being used.

23
Q

conjugate base benzoate is a water-soluble anion = push reaction to the right.

A

Reason for preparing a salt for a non water soluble acidic drug.

24
Q

methyl-amine + water —> ammonium chloride and water

A

methyl-amine will become protonated (added H+)
pKa increases from 1.74 (water) to 10.7 (NH4Cl) because it is a weaker acid

25
Q

why does equilibrium favour the right side?

A

there is a large difference in pKa

26
Q

NaNH2: formation of NH3

A

pKa 5.7 stronger acid (H2O) give a H+ tpo make a weaker base OH- and a weaker acid with a pKa 38

27
Q

H2O donate H+ to any _____ stronger than OH-

A

base

28
Q

to use bases that are stronger than OH- :

A

solvent weaker acid than H2O

29
Q

Examples of weak acids?

A

Hydrocarbons (pKa 50)
Ammonia (pKa 38)

30
Q

Amino acids are _____ compounds;

A

amphoteric

31
Q

amphoteric compounds?

A

act as proton donor and as a proton acceptor (acid/base depending on conditions)

32
Q

ALL amphiprotic substances are _________ - but reverse isn’t true

A

amphoteric

33
Q

Zwitterion;

A

doubly charged form

34
Q

COOH protonates NH2 at pH 7, amino acids having neutral R will exist predominantly as

A

zwitterion (internal salt)

35
Q

ammonium cation > zwitterion net charge is ____> aminocarboxylate anion

A

zero

36
Q

amino acids are _____ compounds - therefore they have no pH existence

A

ionic

37
Q

Isoelectric point pI; pH at which concentration of the zwitterion is max pH which the concentrations of cationic and anionic forms are _____

A

equal

38
Q

pI=pH=

A

pKa1 + pKa2 / 2

39
Q

equation for Ka1?

A

ka1 = [H+] [zwitterion] / [cation]

40
Q

equation for Ka2?

A

Ka2 = [H+] [anion] / [zwitterion]

41
Q

At pI: [cation] = [anion]

A

[cation] = [H+] [zwitterion] / Ka1 = [anion] = Ka2 [zwitterion] / [H+]

42
Q

the isoelectric point (pI)

A

the pH at which a particular molecule carries no net electrical charge

43
Q

Ka1 x Ka2 = [H+]2 so what is the equation for pH

A

pH = pK1 + pKa2 / 2

44
Q

pI: pKa2 (cation) = pKa3 (dipolar ion) / 2

A

examples…

45
Q

amino acids; positively charged at low (acidic) pH, negatively charged at high (basic) pH and zwitterionic at neutral pH

A

implications fro oral absorption and bioavailability of amino acids from diet

46
Q

bioavailability nutrition?

A

the efficiency with which a dietary component is used. systematically through normal metabolic pathways

47
Q

Ionised species (substances with a charge): veryy low lipid solubility

A

unable to permeate through membranes

only non-ionised drug is usally able to cross membranes

48
Q

When using H-H equation =

A

pH = pKa + log [base] / [acid]

49
Q

if we want pH = pKa for drug (weak acid)

A

log [base] / [acid]= log [ionised] / [non-ionised] = 0

50
Q

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

A

pH = pKa + log [ionised] / [non-ionised]

pH = pKa + 1 [ionised / [non-ionised] = 10 10 / 1

which means [ionised] 10 times

51
Q

pH = pKa + log [ ionised] / [non-ionised]
pH = pKa - 1 [ionised] / [ non=ionised] = 0.1

A

1/10 [non-ionised] 10 times

52
Q

calculate fraction of the total dose that is ionised of any pH if the pKa is known

A

fraction [ionised] = 1 / 1 + antilog (pKa - pH)

%ionised = 100 / 1+ antilog (pKa - pH)

53
Q

weak bases

A

fraction [ionised] = 1 / 1 + antilog (pH - pKa)

% ionised = 100 / 1 + antilog (ph - pKa)

PMP-102 (49 decks)

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