AS.3 Acids, Bases And Buffers Flashcards
(40 cards)
Acid
Proton donor
Base
Proton acceptor
Strong acid
Proton donor that dissociates fully
Weak acid
Proton donor that dissociates partially
Alkali
Soluble base that contains OH- ions
Exception -> NH3 + H2O ——> NH4 + + OH-
Acid base pair
Two molecules which interconvert between each other with the loss or gain of a proton
Acid base pair example
HA + B ————-> A- + HB+
HA - acid 1
A- - base 1
Conjugate
B - base 2
HB+ - acid 2
Conjugate
Acid + metal ( ionic equation )
Salt + H2
Na + H+ ———> Na+ + H2
Acid + metal oxide (ionic equation)
Salt + water
MgO + 2H+ ————> Mg2+ + H2O
Acid + alkali (ionic equation)
Salt and water
OH- + H+ ——-> H2O
Acid and carbonate
Salt + water + carbon dioxide
Group I carbonate - soluble
CO3 2- + 2H+ ———-> CO2 + H2O
Group II carbonate - insoluble
MgCO3 + 2H+ ———> Mg2+ + CO2 + H2O
pH equation
-log [H+]
[H+] equation
10^-pH
pH of strong acids
[monobasic acid] = [H+]
pH of weak acids
HA <———-> H+ + A-
Ka = [H+] [A-]/ [HA]
Ka approximates
[HA] eq = [HA] undissociated
Amount of HA considered constant because equilibrium is so far to left
[H+] = [A-]
assume only source of ions is from dissociation of HA
Rearrange Ka to calculate pH
[H+] = square root of (Ka x [HA])
pKa
-log(Ka)
The larger the Ka, the ——- the pKa
The larger the Ka, the more dissociated the acid is, therefore the stronger the weak acid
The larger the Ka, the smaller the pKa
The smaller the pKa, more dissociated the acid is and therefore the stronger the weak acid
pH of pure water
Kw = [H+] [OH-]
Kw at 298K
1 x 10^ -14
Rearranging Kw
Square root of Kw = H+
pH of a strong base
Kw
——— = [ H+]
[NaOH]
Kw = [H+] x [OH-]
[OH-] = [NaOH]
Buffer solution
A solution that resists the change to its pH despite the addition of small amounts of acid or alkali