Acid-base pH III Flashcards by Casi Jones

Polyfunctional acids & bases

Dibasic acid

-2 ionisable groups

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Strong acid

Tribasic

3 ionisable group

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Dicarboxylic acids

extra COOH: e- withdrawing stabilises conjugate base

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Diamines

-NH2 and -NH3- e- withdrawing

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Extra -NH2

not very marked effect on 1st pKa1 reduction

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

diminishes
increases

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

destabilise

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Extra COOH: ______ acidity

increase

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lower the pKa ______ the acidity

higher

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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.

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Extra -NH3+

reducing a lot of the pKa2

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Examples of oxoacids;

HNO3, H2SO4, H3PO4

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

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Na2O is a _______ oxide

basic = reacts with water to make NaOH solution

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SO2 is an ______ oxide

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

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Basic oxides reacts with ____
Acidic oxides reacts with ____

acids
bases

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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)

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Equivalence point; acid-base –>

equal stoichiometric amount

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reaction completed when…

all acid has been converted to its conjugate base

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Speciation curve

look at lecture slides/google

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how to use pKa values

predicts outcome of acid-base interactions

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acid-base interactions

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

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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)