m5 - acids and buffers Flashcards
conjugate acid-base pairings
stay in pairs
acid -> conjugate base
base -> conjugate acid
pH equation on calc
-log [H+]
H+ conc equation on calc
shift
log
—pH
Ka acid dissociation constant
conc products / conc reactants
eg
[CH3COO-] [H+]
————————
[CH3COOH]
assumptions Ka makes
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)
rearranging Ka to find H+ ions
[H+] = square root: Ka x [reactant]
pKa equation
-log Ka
Kw equation (ionic product of water)
[H+] [OH-]
buffer def
resists changes in pH when small quantities of an acid or alkali are added to it
what’s in a buffer solution
a weak acid eg CH3COOH
salt of the weak acid eg CH3COONa
partial: CH3COOH -><- CH3COO- + H+
complete: CH3COONa -> CH3COO- + Na+
buffer Ka equation
[salt] [H+]
—————— = Ka
[acid]
buffers in the body : carbonic acid - hydrogencarbonate system
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
titration curves - equivalence point to work out Ka
at half the equivalence point pH = pKa
pKa = -logKa
what indicator to use if vertical part of a titration curve is pH 3-5
methyl orange (red=>yellow)
what indicator to use if vertical part of a titration curve is pH 5-8
litmus (red=>blue)
what indicator to use if vertical part of a titration curve is pH 8-10
phenolphthalein (colourless=>pink)
Ka > 1 significance
equilibrium on the right, more products than reactants
so quite a strong weak acid
finding pH in strong acids
mole ratio is 1:1 HA —> H+. + A-
[HA] = [H+]
therefore pH = - log [HA]
finding pH in a strong base
1:1 ratio
MOH —> M+ + OH-
[MOH] = [OH-]
rearrange and use Kw to find [H+]
pH = - log [H+]
finding pH in weak acid
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
adding alkali to a buffer
HA -><- H + and A -
OH- reacts with H+ -> h2o
so eq shifts to the right , so more dissociation occurs and H+ replaced, pH restored
adding acid to buffer
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
what’s the point of a buffer with excess weak acid and strong alkali
these react together to form the salt of the weak acid
excess weak acid available and buffer formed
finding pH of buffer
[H+] ≠ [A-] so…
assume A- conc = salt conc
Ka = [H+] [salt]
——————
[weak acid]
so
[H+] = Ka x [weak acid]
————————
[salt]
