acid and bases (2) Flashcards
how to use ion product of water
shows relationship between [H3O+] or [OH-] in aq
know one conc of one species and use Kw to find other
Kw equation - for water - ions
[H3O+][OH-]
= 1x10(-14)
Kw equation - acid and bases
Kw = Ka x Kb = 1x10(-14)
Kb
weak base
calculate Ka for its conjugate acid
Ka
weak acid
calculate Kb for its conjugate base
pH and conc of acid
[H+] varies in aq therefore difficult to compare
use pH=-log[H+]
weak acid and [H+] - dissociate partially
e.g. ethanoic acid - 0.1M
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+]
Henderson-Hasselbalch equation
pH = pka + log([A-]/[HA])
contribution of water
OH- and H3O+ ions - inversely proportional
acidic - water
[H3O+] - relatively high
[OH-] - relatively low
basic - water
[H3O+] - relatively low
[OH-] - relatively high
general dissociation of acid equation
HA + H2O A- + H3O+
once dissociated uncharged HA -> charged A-
stronger acid
ionised in aq solution therefore partition into aq phase
weaken acid
not ionised therefore likely to partition into membrane
effect of local environment - pH decrease
decrease in dissociation of weak acid - increase [H+]
therefore acid system HAA-+H+ respond increase [H+] and shifts left