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# Ocr A: weak acid approximations watch

1. Could you please explain the approximations of the weak acids

My text book says:

1) HA dissociates to produce equilibrium concentrations of H+ and A- that are equal. There will also be a very small concentration of H+ from the dissociated of water but this will be extremely small and can be neglected compared with the H+ concentration from the acid

[H+]=[A-]

What does this paragraph mean and how does the equation above relate

2) the equilibrium conecentreration of Ha is smaller than the undicsocaited concentration

[HA]eqm = [HA]start - [H+]eqm

As the dissociation of weak acids is small, you can assume that [HA] START >> [H+] and you can neglect any decrease in the concentration of HA from disociation

[HA]eqm = [HA]start

What???
2. (Original post by Frank Peters)
Could you please explain the approximations of the weak acids
Before I begin, have you been taught how to calculate the pH of strong bases?
3. (Original post by Pigster)
Before I begin, have you been taught how to calculate the pH of strong bases?
yes

pH = -log(H+)
4. (Original post by Frank Peters)
yes

pH = -log(H+)
That's for strong acids, not bases. Have you learnt about Kw?

(also I hope you write square brackets, not round ones. For the record the square brackets on a standard keyboard are the right of the p key)
5. (Original post by Pigster)
That's for strong acids, not bases. Have you learnt about Kw?

(also I hope you write square brackets, not round ones. For the record the square brackets on a standard keyboard are the right of the p key)
aaa, sorry, I remember strong bases: use Kw equation and then use pH=-log(H+)

Kw = [H+] [OH-]
6. (Original post by Frank Peters)
1) HA dissociates to produce equilibrium concentrations of H+ and A- that are equal. There will also be a very small concentration of H+ from the dissociated of water but this will be extremely small and can be neglected compared with the H+ concentration from the acid
When you dissolve HA in water, there are two sources of H+, from the dissociation of HA AND from the dissociation of H2O.

The 1st assumption is that [H+](from HA) >> [H+](from H2O), i.e. you can ignore the H+ from the H2O.

(Original post by Frank Peters)
2) the equilibrium conecentreration of Ha is smaller than the undicsocaited concentration

[HA]eqm = [HA]start - [H+]eqm

As the dissociation of weak acids is small, you can assume that [HA] START >> [H+] and you can neglect any decrease in the concentration of HA from disociation

[HA]eqm = [HA]start

What???
The 2nd assumption is basically that the dissociation of HA is sooo small that you can assume (to 3sf) that NONE of the weak acid dissociates, i.e. whatever the concentration of the weak acid is before any of it dissolves is (again to 3sf) the same as the concentration as it is after some it dissociates.
7. (Original post by Pigster)
When you dissolve HA in water, there are two sources of H+, from the dissociation of HA AND from the dissociation of H2O.

The 1st assumption is that [H+](from HA) >> [H+](from H2O), i.e. you can ignore the H+ from the H2O.

The 2nd assumption is basically that the dissociation of HA is sooo small that you can assume (to 3sf) that NONE of the weak acid dissociates, i.e. whatever the concentration of the weak acid is before any of it dissolves is (again to 3sf) the same as the concentration as it is after some it dissociates.
I still dont get the first assumption. like how would H20 be involved. Could you possibly give me a full weak acid equation with h20
8. (Original post by Frank Peters)
I still dont get the first assumption. like how would H20 be involved. Could you possibly give me a full weak acid equation with h20
also, why would it mean that this assumption means
[H+]eqm = [A-]eqm
9. (Original post by Frank Peters)
I still dont get the first assumption. like how would H20 be involved. Could you possibly give me a full weak acid equation with h20
Essentially, OCR A use moderately weak acids, i.e. those that dissociate a bit, but not too much.

Water is effectively a super weak acid. Let us say that 1 in 10 000 000 molecules of water turns into H+ and OH-.

If you add the super weak acid, HA, which also has 1 molecule in 10 000 000 dissociate... if you measure the pH of this HA solution, then where are the H+ ions coming from? (the ones you're detecting when you're measuring pH). From both the HA AND from the water.

The first assumption is that ALL of the H+ come from the HA, rather from the water. i.e. the weak acid isn't VERY weak.

The second assumption is that the acid isn't especially strong.

Ka = [H+][A-] / [HA]

If the acid were a relatively strong acid, then when it dissolves in water the [HA] would decrease (as it is dissociating to H+ and A-). This would mean you'd have to use quadratics to calculate pH (like we used to have to). But if HA is a relatively weak acid, then [HA] doesn't change by much when it dissolves.
10. (Original post by Pigster)
Essentially, OCR A use moderately weak acids, i.e. those that dissociate a bit, but not too much.

Water is effectively a super weak acid. Let us say that 1 in 10 000 000 molecules of water turns into H+ and OH-.

If you add the super weak acid, HA, which also has 1 molecule in 10 000 000 dissociate... if you measure the pH of this HA solution, then where are the H+ ions coming from? (the ones you're detecting when you're measuring pH). From both the HA AND from the water.

The first assumption is that ALL of the H+ come from the HA, rather from the water. i.e. the weak acid isn't VERY weak.

The second assumption is that the acid isn't especially strong.

Ka = [H+][A-] / [HA]

If the acid were a relatively strong acid, then when it dissolves in water the [HA] would decrease (as it is dissociating to H+ and A-). This would mean you'd have to use quadratics to calculate pH (like we used to have to). But if HA is a relatively weak acid, then [HA] doesn't change by much when it dissolves.

ok, so why would the book then tell me that [H+]eqm =[A-]eqm
11. (Original post by Frank Peters)
ok, so why would the book then tell me that [H+]eqm =[A-]eqm
That's the 1st assumption.

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