Buffers a level chemistry Watch

Bertybassett
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Hello, for question 5b on this question I am a bit confused. I got the correct answer by just using the moles (instead of the concentration) because no volume was provided? How is this possible as I thought the ka equation required the concentrations? Many thanks https://pmt.physicsandmathstutor.com...n%201%20QP.pdf
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BobbJo
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HO2CCO2H -> HO2CCOO- + H+
Ka = [H+][A-]/[HA]
H+, A- and HA are all in the same volume. Let the volume of the solution be V
[A-] = n(A-)/V
[HA] = n(HA)/V

hence replace the above in Ka,
 K_a = \dfrac{ [H^+] \times \frac{n(A^-)}{V} }{ \frac{n(HA)}{V}}
so V cancels
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Bertybassett
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is there a way of knowing when and when you cant do this? surely this works for all ka calculations if all the substances are in the same volume, in fact when wouldn't the substances be in the same volume?
(Original post by BobbJo)
HO2CCO2H -> HO2CCOO- + H+
Ka = [H+][A-]/[HA]
H+, A- and HA are all in the same volume. Let the volume of the solution be V
[A-] = n(A-)/V
[HA] = n(HA)/V

hence replace the above in Ka,
 K_a = \dfrac{ [H^+] \times \frac{n(A^-)}{V} }{ \frac{n(HA)}{V}}
so V cancels
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BobbJo
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(Original post by Bertybassett)
is there a way of knowing when and when you cant do this? surely this works for all ka calculations if all the substances are in the same volume, in fact when wouldn't the substances be in the same volume?
They are in the same volume meaning they are in the same solution.. same containing vessel i.e same beaker/conical flask/etc

The substances would not be in the same volume if they are not in the same beaker/conical flask/etc

You can know when and when you can't do this by replacing into the expression for Ka and seeing if V cancels
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Bertybassett
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but how do you know if v does cancel? I don't really understand how v cancels in your previous post
(Original post by BobbJo)
They are in the same volume meaning they are in the same solution.. same containing vessel i.e same beaker/conical flask/etc

The substances would not be in the same volume if they are not in the same beaker/conical flask/etc

You can know when and when you can't do this by replacing into the expression for Ka and seeing if V cancels
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BobbJo
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(Original post by Bertybassett)
but how do you know if v does cancel? I don't really understand how v cancels in your previous post
 K_a = \dfrac{ [H^+] \times \frac{n(A^-)}{V} }{ \frac{n(HA)}{V}}

consider the ratio [A-]/[HA]
[A-]/[HA]=  \dfrac{n(A^-)}{V} \times \dfrac{V}{n(HA)}  = \dfrac{n(A^-)}{n(HA)}

so  K_a = \dfrac{ [H^+] \times n(A^-)}{ n(HA)}
Last edited by BobbJo; 1 week ago
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Bertybassett
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(Original post by BobbJo)
 K_a = \dfrac{ [H^+] \times \frac{n(A^-)}{V} }{ \frac{n(HA)}{V}}

consider the ratio [A-]/[HA]
[A-]/[HA]=  \dfrac{n(A^-)}{V} \times \dfrac{V}{n(HA)}  = \dfrac{n(A^-)}{n(HA)}

so  K_a = \dfrac{ [H^+] \times n(A^-)}{ n(HA)}
thanks I understand that now. Also for this question 2b on this paper, I am a bit stuck on how to work this out. For example for 2bi, I get that h2o is a base as it accepts the proton, but how is ch3coo- a base too? Surely it has donated a proton so it is an acid? Im completely stuck on it. thanks https://pmt.physicsandmathstutor.com...n%202%20QP.pdf
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charco
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(Original post by Bertybassett)
thanks I understand that now. Also for this question 2b on this paper, I am a bit stuck on how to work this out. For example for 2bi, I get that h2o is a base as it accepts the proton, but how is ch3coo- a base too? Surely it has donated a proton so it is an acid? Im completely stuck on it. thanks https://pmt.physicsandmathstutor.com...n%202%20QP.pdf
In Bronsted Lowry theory a substance that accepts a hydrogen ion is behaving as a base and a substance that donates a hydrogen ion is behaving as an acid.

In any acid-base equation there will be an acid on the left and its conjugate base on the right AND a base on the left and its conjugate acid on the right. In other words, there are two acid-base pairs.

ACID + BASE --> CONJUGATE BASE + CONJUGATE ACID
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Bertybassett
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(Original post by charco)
In Bronsted Lowry theory a substance that accepts a hydrogen ion is behaving as a base and a substance that donates a hydrogen ion is behaving as an acid.

In any acid-base equation there will be an acid on the left and its conjugate base on the right AND a base on the left and its conjugate acid on the right. In other words, there are two acid-base pairs.

ACID + BASE --> CONJUGATE BASE + CONJUGATE ACID
right thanks, but still how do you know which is which. So if the acid is say ethanoic acid on the lhs, why would the enthanoate ion be the base?
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BobbJo
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(Original post by Bertybassett)
right thanks, but still how do you know which is which. So if the acid is say ethanoic acid on the lhs, why would the enthanoate ion be the base?
CH3COOH + H2O <-> CH3COO- + H3O+
forward reaction: CH3COOH is donating a proton to H2O, forming CH3COO- and H3O+
Bronsted-Lowry theory: Acids are proton donors. Bases are proton acceptors.
CH3COOH donated a proton so it is an acid.
H2O accepted a proton so it is a base.

Reverse reaction: CH3COO- accepts a proton from H3O+, forming CH3COOH and H2O.
CH3COO- is accepting a proton, so it is a base
H3O+ is donating a proton, so it is an acid

CH3COOH and CH3COO- are conjugate acid-base pairs.
H2O and H3O+ are conjugate acid-base pairs.
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