✅12 - Acid Base Equilibria
No description yet. This curated deck is ready for daily study.
Card preview (24)
What is a Bronsted-Lowry acid?
A proton donor
What is a Bronsted-Lowry base?
A proton acceptor
What is a conjugate acid?
A species formed by the reception of a proton (H+) by a base in other words, it is a base with a hydrogen ion added to it.
What is a conjugate base?
What is left after an acid has donated a proton in a chemical reaction.
What is a diprotic/dibasic acid?
One which can donate two protons, such as H2SO4
What is a diprotic/diacidic base?
One which can accept two protons, such as CO3 2-
What are amphoteric substances?
One which can act either as an acid or a base, such as water
What is a strong acid?
One which is almost completely dissociated in aqueous solution
How are strong acids represented?
With a single arrow in the dissociation equation, not reversible
What is a weak acid?
One that is only partially dissociated in aqueous solution
How are weak acids represented?
With a reversible arrow in the dissociation equation
How are hydrogen ion concentration and and concentration linked for strong acids?
Directly, so HCl of 0.100 moldm-3 will have an H+ concentration of 0.100 moldm-3
How would you calculate pH from H+ concentration?
pH = -log [H+]
How would you calculate [H+] from pH?
[H+] = 10^-pH
What is Ka?
The acid dissociation constant
What is the equation to work out Ka of a weak acid?
[H+] x [A-] _______ [HA]
What can the expression for Ka be simplified to?
[H+]^2 _____ [HA]
How can you work out [H+] for a weak acid from Ka?
By rearranging the equation to find a value for [H+]^2, then square rooting the value.
How would you work out pKa?
-log Ka
The larger the value of Ka…
…the stronger the acid
The larger the value of pKa…
…the weaker the acid
How would you calculate the pH of a dibasic acid?
Ka HSO4 = 0.0100 In 0.500 moldm-3 solution of H2SO4, contribution to [H+] from H2SO4 will be 0.500moldm-3. If we assume that contribution to [H+] from HSO4- is (x)moldm-3: Ka(HSO4-) = 0.0100 = (0.500- x) x/0.500 Solving this equation gives x = 0.0098 So total [H+] = 0.05098 with a pH of 0.293
What is Kw?
The ionic product of water
What is the equation to calculate Kw?
[H+] x [OH-] _________ [H2O]