The pH of weak acids Flashcards
What is a weak acid in terms of HA?
HA partially dissociates.
What is a strong acid in terms of HA?
HA completely dissociates.
How do you represent a weak acid in an equation using HA?
HA (aq) = H+ (aq) + A- (aq)
- Where = is an equilibrium sign.
What 2 things does the H+ concentration depend upon? (2)?
- The concentration of the acid, [HA].
- The acid dissociation constant, Ka.
What happens at equilibrium when [HA] dissociates?
H+ and A- ions are formed in equal quantities:
- Where A- is the negative ion.
What is the equation for calculating Ka using equilibrium concentrations?
How can this equation be simplified using 2 approximations? (2)?
Therefore, what is the simplified equation for Ka using equilibrium concentrations?
[H+ (aq) at =] / [HA (aq)] at =]
=
[H+ (aq)] at =] x [A- (aq)] at =] /
[HA (aq)] at start] x [H+ (aq) at =].
- Approximation 1:
HA dissociates to produce equilibrium concentrations of H+ (aq) and A- (aq) that are equal. There will be some H+ (aq) from the water but this can be neglected compared to the acid:
([H+ (aq)] at equilibrium) = ([A- (aq)] at equilibrium).
- Approximation 2:
The equilibrium concentration of HA is smaller than the undissociated concentration:
[HA (aq) at equiibrium] = ([HA (aq) at start] - [H+ (aq) at equilibrium).
As the dissociation of weak acids are small, you can assume that [HA] at the start is bigger than [H+] and you can neglect any decrease the concentration of HA from dissociation.
- Therefore by applying the two approximations, the expression can be greatly simplified:
Ka = [H+ (aq)]^2 / [HA (aq)].
What is the equation for working out H+ of a weak acid using the Ka equation?
[H+ (aq)] = ROOT Ka x [HA (aq)]
What is the process for working out pH of a weak acid using the Ka equation?
- ‘Calculate the pH of 0.0245 moldm-3 ethanoic acid, CH3COOH, at 25C, where Ka = 1.7 x 10^-5 moldm-3?’.
- Calculate [H+ (aq)] from Ka and [HA (aq)]:
- CH3COOH is a weak acid and partially dissociates.
- CH3COOH (aq) = H+ (aq) + CH3COO- (aq).
Ka = [H+ (aq)]^2 / [CH3COOH (aq)].
[H+ (aq)]^2 = Ka x [CH3COOH (aq)].
[H+ (aq) = ROOT (Ka x [CH3COOH (aq)] = ROOT (0.0245 x (1.7 x 10^-5)).
= 6.45 x 10^-4 moldm-3.
- Use your calculator to find the pH:
pH = -log[H+ (aq)] = -log (6.45 x 10^-4) = 3.19.
How do you determine Ka of a weak acid experimentally? (2)?
- Prepare a standard solution of the weak acid of known concentration.
- Measure the pH of the standard solution using a pH meter.
What is the process for working out Ka for a weak acid?
‘The pH of 0.065moldm-3 propanoic acid, CH3CH2COOH, is 3.04. Calculate Ka? (2)?
- Use your calculator to find [H+ (aq)]:
[H+ (aq)] = 10^-pH = 10^-3.04 = 9.12 x 10^-4 moldm-3.
- Calculate Ka from [H+ (aq)] and [HA (aq)]:
Ka = [H+ (aq)]^2 / [CH3CH2COOH (aq)].
Ka = (9.12 x 10^-4)^2 / 0.065
Ka = 1.28 x 10^-5 moldm-3.
What is the first approximation of calculations for weak acids?
What does it assume?
Is it justified?
When does this approximation break down?
([H+ (aq)] at eq). = ([A- (aq)] at eq.).
It assumes that the dissociation of water is negligible.
If the pH is less than 6, then H+ concentration from the dissociation of water will be significant compared with the dissociation of a weak acid.
It breaks down for very weak acids or very dilute solutions.
What is the second approximation of calculations for weak acids?
What does it assume?
Is it justified?
When does this approximation break down?
([HA] at eq.) = ([HA] at start) - ([H+ et eq.) approximates to ([HA] at eq. = [HA] at start).
It assumes that the concentration of acid is much greater than the H+ concentration at equilibrium.
This approximation is not justified for stronger weak acids with Ka less than 10^-2 moldm-3 and for very dilute solutions.
It will hold for weak acids with small Ka values. It breaks down when [H+] becomes significant and there is a real difference between [HA] at eq. and ([HA] at start - [H+] at eq).
