acid and bases (2) Flashcards

1
Q

how to use ion product of water

A

shows relationship between [H3O+] or [OH-] in aq

know one conc of one species and use Kw to find other

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2
Q

Kw equation - for water - ions

A

[H3O+][OH-]

= 1x10(-14)

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3
Q

Kw equation - acid and bases

A

Kw = Ka x Kb = 1x10(-14)

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4
Q

Kb

A

weak base

calculate Ka for its conjugate acid

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5
Q

Ka

A

weak acid

calculate Kb for its conjugate base

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6
Q

pH and conc of acid

A

[H+] varies in aq therefore difficult to compare

use pH=-log[H+]

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7
Q

weak acid and [H+] - dissociate partially

e.g. ethanoic acid - 0.1M

A

1- make equation of acid with water
2- have initial conc of each molecule
3- have conc of molecules at equilibrium
4- Ka=[CH3COO-][H3O-] / [CH3COOH] = 1.8x10(-5)
5- solve x - assuming a is v. small - change in concentration [H+]
6- pH = -log[H+]

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8
Q

Henderson-Hasselbalch equation

A

pH = pka + log([A-]/[HA])

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9
Q

contribution of water

A

OH- and H3O+ ions - inversely proportional

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10
Q

acidic - water

A

[H3O+] - relatively high

[OH-] - relatively low

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11
Q

basic - water

A

[H3O+] - relatively low

[OH-] - relatively high

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12
Q

general dissociation of acid equation

A

HA + H2O A- + H3O+

once dissociated uncharged HA -> charged A-

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13
Q

stronger acid

A

ionised in aq solution therefore partition into aq phase

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14
Q

weaken acid

A

not ionised therefore likely to partition into membrane

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15
Q

effect of local environment - pH decrease

A

decrease in dissociation of weak acid - increase [H+]

therefore acid system HAA-+H+ respond increase [H+] and shifts left

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16
Q

effect of local environment - pH increase

A

increase in dissociation of weak acid

17
Q

acid in pure water - a equilibrium

A

is set up = conjugated base etc

know amount of acid dissociated

18
Q

pKa corresponds to pH at 50% of acid dissociated

A

pHpKa - >50% dissociated

19
Q

[A-]

A

conjugated base

20
Q

[HA]

A

conc of acid

21
Q

equation at pure water

A

[H3O+] = [OH-]