m5 - acids and buffers

Question 1

conjugate acid-base pairings

Answer 1

stay in pairs
acid -> conjugate base
base -> conjugate acid

Question 2

pH equation on calc

Answer 2

-log [H+]

Question 3

H+ conc equation on calc

Answer 3

shift
log
—pH

Question 4

Ka acid dissociation constant

Answer 4

conc products / conc reactants

eg
[CH3COO-] [H+]
————————
[CH3COOH]

Question 5

assumptions Ka makes

Answer 5

conc products are equal because mole ratio 1:1, so can write products as [H+]^2

conc of weak acid is the same at the start and at equilibrium as it doesn’t dissociate fully (if the weak acid has dissociated more than 5% then we can’t use the above approximation)

Question 6

rearranging Ka to find H+ ions

Answer 6

[H+] = square root: Ka x [reactant]

Question 7

pKa equation

Answer 7

-log Ka

Question 8

Kw equation (ionic product of water)

Answer 8

[H+] [OH-]

Question 9

buffer def

Answer 9

resists changes in pH when small quantities of an acid or alkali are added to it

Question 10

what’s in a buffer solution

Answer 10

a weak acid eg CH3COOH
salt of the weak acid eg CH3COONa

partial: CH3COOH -><- CH3COO- + H+
complete: CH3COONa -> CH3COO- + Na+

Question 11

buffer Ka equation

Answer 11

[salt] [H+]
—————— = Ka
[acid]

Question 12

buffers in the body : carbonic acid - hydrogencarbonate system

Answer 12

H2CO3 -><- HCO3- + H+

if OH- added, they react with H+, equilibrium shifts to right to regenerate more H+
if H+ added, equilibrium shifts to left to remove them

Question 13

titration curves - equivalence point to work out Ka

Answer 13

at half the equivalence point pH = pKa

pKa = -logKa

Question 14

what indicator to use if vertical part of a titration curve is pH 3-5

Answer 14

methyl orange (red=>yellow)

Question 15

what indicator to use if vertical part of a titration curve is pH 5-8

Answer 15

litmus (red=>blue)

Question 16

what indicator to use if vertical part of a titration curve is pH 8-10

Answer 16

phenolphthalein (colourless=>pink)

Question 17

Ka > 1 significance

Answer 17

equilibrium on the right, more products than reactants

so quite a strong weak acid

Question 18

finding pH in strong acids

Answer 18

mole ratio is 1:1 HA —> H+. + A-
[HA] = [H+]

therefore pH = - log [HA]

Question 19

finding pH in a strong base

Answer 19

1:1 ratio
MOH —> M+ + OH-
[MOH] = [OH-]

rearrange and use Kw to find [H+]
pH = - log [H+]

Question 20

finding pH in weak acid

Answer 20

HA -><- H+ + A-
Ka = [H+] [A-]
————
[HA]
which is also equal to
Ka = [H+]^2
————
[HA]

[H+]^2 = Ka x [HA]
square root
find pH

Question 21

adding alkali to a buffer

Answer 21

HA -><- H + and A -

OH- reacts with H+ -> h2o
so eq shifts to the right , so more dissociation occurs and H+ replaced, pH restored

Question 22

adding acid to buffer

Answer 22

NaA —> Na+ and A-

eq shifts to the left, so H+ and A- react to form HA
this removes the H+ that was added and restores pH

Question 23

what’s the point of a buffer with excess weak acid and strong alkali

Answer 23

these react together to form the salt of the weak acid
excess weak acid available and buffer formed

Question 24

finding pH of buffer

Answer 24

[H+] ≠ [A-] so…
assume A- conc = salt conc

Ka = [H+] [salt]
——————
[weak acid]
so

[H+] = Ka x [weak acid]
————————
[salt]