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Joynzer
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Hi, I have a couple of questions about the manner in which questions regarding "finding the pH of buffer" are worded within exams. I'm getting really confused and wondered if anyone could offer some advice?
For example, in the OCR June 2012 F325 paper, the question states:
The student adds 50.0cm^3 of 0.250moldm^-3 butanoic acid to 50.0cm^3 of 0.0500moldm^-3 sodium hydroxide. Calculate the pH of the buffer solution. (You are then given a value of Ka).
Now, I appreciate that to answer this question the initial number of moles need to be established, and following that, the moles that actually reacted. However, is this due to the fact that we are reacting an acid and a base together to form the salt (which then dissociates to form the conjugate base), and not initially using a mixture of the acid and it's salt.
So in this question we have the acid + base, forming the salt. The salt then disassociates to form the conjugate base.
However if we started with the acid, and a known volume and concentration of SODIUM BUTANOATE (the salt) we would not need to do this stoichiometry part because the acid and base don't actually react to form the salt?
Sorry if that is really confusing, any help would be appreciated.
For example, in the OCR June 2012 F325 paper, the question states:
The student adds 50.0cm^3 of 0.250moldm^-3 butanoic acid to 50.0cm^3 of 0.0500moldm^-3 sodium hydroxide. Calculate the pH of the buffer solution. (You are then given a value of Ka).
Now, I appreciate that to answer this question the initial number of moles need to be established, and following that, the moles that actually reacted. However, is this due to the fact that we are reacting an acid and a base together to form the salt (which then dissociates to form the conjugate base), and not initially using a mixture of the acid and it's salt.
So in this question we have the acid + base, forming the salt. The salt then disassociates to form the conjugate base.
However if we started with the acid, and a known volume and concentration of SODIUM BUTANOATE (the salt) we would not need to do this stoichiometry part because the acid and base don't actually react to form the salt?
Sorry if that is really confusing, any help would be appreciated.
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langlitz
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#2
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#2
(Original post by Joynzer)
Hi, I have a couple of questions about the manner in which questions regarding "finding the pH of buffer" are worded within exams. I'm getting really confused and wondered if anyone could offer some advice?
For example, in the OCR June 2012 F325 paper, the question states:
The student adds 50.0cm^3 of 0.250moldm^-3 butanoic acid to 50.0cm^3 of 0.0500moldm^-3 sodium hydroxide. Calculate the pH of the buffer solution. (You are then given a value of Ka).
Now, I appreciate that to answer this question the initial number of moles need to be established, and following that, the moles that actually reacted. However, is this due to the fact that we are reacting an acid and a base together to form the salt (which then dissociates to form the conjugate base), and not initially using a mixture of the acid and it's salt.
So in this question we have the acid + base, forming the salt. The salt then disassociates to form the conjugate base.
However if we started with the acid, and a known volume and concentration of SODIUM BUTANOATE (the salt) we would not need to do this stoichiometry part because the acid and base don't actually react to form the salt?
Sorry if that is really confusing, any help would be appreciated.
Hi, I have a couple of questions about the manner in which questions regarding "finding the pH of buffer" are worded within exams. I'm getting really confused and wondered if anyone could offer some advice?
For example, in the OCR June 2012 F325 paper, the question states:
The student adds 50.0cm^3 of 0.250moldm^-3 butanoic acid to 50.0cm^3 of 0.0500moldm^-3 sodium hydroxide. Calculate the pH of the buffer solution. (You are then given a value of Ka).
Now, I appreciate that to answer this question the initial number of moles need to be established, and following that, the moles that actually reacted. However, is this due to the fact that we are reacting an acid and a base together to form the salt (which then dissociates to form the conjugate base), and not initially using a mixture of the acid and it's salt.
So in this question we have the acid + base, forming the salt. The salt then disassociates to form the conjugate base.
However if we started with the acid, and a known volume and concentration of SODIUM BUTANOATE (the salt) we would not need to do this stoichiometry part because the acid and base don't actually react to form the salt?
Sorry if that is really confusing, any help would be appreciated.
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Borek
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#3
Note: technically you could argue that after neutralization reaction goes to completion, it is followed by either acid dissociation, or base hydrolysis, and equilibrium concentrations are different from those calculated from the neutralization - and you would be right. But this is quite a reasonable approximation, see http://www.chembuddy.com/?left=buffe...with-ICE-table (note: it doesn't matter how the solution is prepared, by mixing acid with NaOH, or acid with sodium salt).
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bahiadabo
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#4
(Original post by Joynzer)
Hi, I have a couple of questions about the manner in which questions regarding "finding the pH of buffer" are worded within exams. I'm getting really confused and wondered if anyone could offer some advice?
For example, in the OCR June 2012 F325 paper, the question states:
The student adds 50.0cm^3 of 0.250moldm^-3 butanoic acid to 50.0cm^3 of 0.0500moldm^-3 sodium hydroxide. Calculate the pH of the buffer solution. (You are then given a value of Ka).
Now, I appreciate that to answer this question the initial number of moles need to be established, and following that, the moles that actually reacted. However, is this due to the fact that we are reacting an acid and a base together to form the salt (which then dissociates to form the conjugate base), and not initially using a mixture of the acid and it's salt.
So in this question we have the acid + base, forming the salt. The salt then disassociates to form the conjugate base.
However if we started with the acid, and a known volume and concentration of SODIUM BUTANOATE (the salt) we would not need to do this stoichiometry part because the acid and base don't actually react to form the salt?
Sorry if that is really confusing, any help would be appreciated.
Hi, I have a couple of questions about the manner in which questions regarding "finding the pH of buffer" are worded within exams. I'm getting really confused and wondered if anyone could offer some advice?
For example, in the OCR June 2012 F325 paper, the question states:
The student adds 50.0cm^3 of 0.250moldm^-3 butanoic acid to 50.0cm^3 of 0.0500moldm^-3 sodium hydroxide. Calculate the pH of the buffer solution. (You are then given a value of Ka).
Now, I appreciate that to answer this question the initial number of moles need to be established, and following that, the moles that actually reacted. However, is this due to the fact that we are reacting an acid and a base together to form the salt (which then dissociates to form the conjugate base), and not initially using a mixture of the acid and it's salt.
So in this question we have the acid + base, forming the salt. The salt then disassociates to form the conjugate base.
However if we started with the acid, and a known volume and concentration of SODIUM BUTANOATE (the salt) we would not need to do this stoichiometry part because the acid and base don't actually react to form the salt?
Sorry if that is really confusing, any help would be appreciated.
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Borek
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#5
The most common source of confusion lies in the fact that buffer solution contains a weak acid and its CONJUGATE base. NaOH is not the base we are talking about, it is used to PRODUCE the conjugate base that will be part of the buffer.
Does it help?
Does it help?
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Joynzer
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#6
So if the strong base was in excess then there would be no buffer right? If however, we had a situation where the buffer was being formed from say butanoic acid and sodium bethanoate (and the sodium bethanoate was in excess) a buffer solution could still be formed?
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niceguy95
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#7
I got 4.22 and this is how i worked it.
1) First right out Equation
Butanoic Acid + NaOh <---> Conjugate Base (salt) + H20 (components)
2) For each Compenent we have a 1:1 ratio
3)We then work out Initial Moles of the given
So moles of Butanoic acid is 0.0125
And moles of NaOH is 0.0025
Current Moles of the Conjugate base (Salt), we assume is 0
4)We then use the smallest moles, in this case the moles for NaOH, we use this to work the Equillibirum moles
How? initial moles +/- Smallest Moles worked out
Equillibrium Moles:
NaOH = 0.0025-0.00025= 0
butanoic acid = 0.0125-0.0025= 0.01
Conjugate base (Salt) = 0 + 0.00025= 0.0025
If your wondering why i added for the salt and subtracted for the others is because, across the equilibrium sign, the sign changes and vice versa, but the smallest moles remains the same.
5) Calculating pH
-logKa + log(Conjugate base/Weak Acid)
We use the equilibrium moles for above as well as the given Ka which is 1.51x10-5
-log(1.51x10-5) + log(0.0025/0.01)
pH= 4.22
EDIT: I hope i got right lol
1) First right out Equation
Butanoic Acid + NaOh <---> Conjugate Base (salt) + H20 (components)
2) For each Compenent we have a 1:1 ratio
3)We then work out Initial Moles of the given
So moles of Butanoic acid is 0.0125
And moles of NaOH is 0.0025
Current Moles of the Conjugate base (Salt), we assume is 0
4)We then use the smallest moles, in this case the moles for NaOH, we use this to work the Equillibirum moles
How? initial moles +/- Smallest Moles worked out
Equillibrium Moles:
NaOH = 0.0025-0.00025= 0
butanoic acid = 0.0125-0.0025= 0.01
Conjugate base (Salt) = 0 + 0.00025= 0.0025
If your wondering why i added for the salt and subtracted for the others is because, across the equilibrium sign, the sign changes and vice versa, but the smallest moles remains the same.
5) Calculating pH
-logKa + log(Conjugate base/Weak Acid)
We use the equilibrium moles for above as well as the given Ka which is 1.51x10-5
-log(1.51x10-5) + log(0.0025/0.01)
pH= 4.22
EDIT: I hope i got right lol
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langlitz
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#8
(Original post by niceguy95)
I got 4.22 and this is how i worked it.
1) First right out Equation
Butanoic Acid + NaOh <---> Conjugate Base (salt) + H20 (components)
2) For each Compenent we have a 1:1 ratio
3)We then work out Initial Moles of the given
So moles of Butanoic acid is 0.0125
And moles of NaOH is 0.0025
Current Moles of the Conjugate base (Salt), we assume is 0
4)We then use the smallest moles, in this case the moles for NaOH, we use this to work the Equillibirum moles
How? initial moles +/- Smallest Moles worked out
Equillibrium Moles:
NaOH = 0.0025-0.00025= 0
butanoic acid = 0.0125-0.0025= 0.01
Conjugate base (Salt) = 0 + 0.00025= 0.0025
If your wondering why i added for the salt and subtracted for the others is because, across the equilibrium sign, the sign changes and vice versa, but the smallest moles remains the same.
5) Calculating pH
-logKa + log(Conjugate base/Weak Acid)
We use the equilibrium moles for above as well as the given Ka which is 1.51x10-5
-log(1.51x10-5) + log(0.0025/0.01)
pH= 4.22
EDIT: I hope i got right lol
I got 4.22 and this is how i worked it.
1) First right out Equation
Butanoic Acid + NaOh <---> Conjugate Base (salt) + H20 (components)
2) For each Compenent we have a 1:1 ratio
3)We then work out Initial Moles of the given
So moles of Butanoic acid is 0.0125
And moles of NaOH is 0.0025
Current Moles of the Conjugate base (Salt), we assume is 0
4)We then use the smallest moles, in this case the moles for NaOH, we use this to work the Equillibirum moles
How? initial moles +/- Smallest Moles worked out
Equillibrium Moles:
NaOH = 0.0025-0.00025= 0
butanoic acid = 0.0125-0.0025= 0.01
Conjugate base (Salt) = 0 + 0.00025= 0.0025
If your wondering why i added for the salt and subtracted for the others is because, across the equilibrium sign, the sign changes and vice versa, but the smallest moles remains the same.
5) Calculating pH
-logKa + log(Conjugate base/Weak Acid)
We use the equilibrium moles for above as well as the given Ka which is 1.51x10-5
-log(1.51x10-5) + log(0.0025/0.01)
pH= 4.22
EDIT: I hope i got right lol
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bahiadabo
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#9
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#9
(Original post by niceguy95)
I got 4.22 and this is how i worked it.
1) First right out Equation
Butanoic Acid + NaOh <---> Conjugate Base (salt) + H20 (components)
2) For each Compenent we have a 1:1 ratio
3)We then work out Initial Moles of the given
So moles of Butanoic acid is 0.0125
And moles of NaOH is 0.0025
Current Moles of the Conjugate base (Salt), we assume is 0
4)We then use the smallest moles, in this case the moles for NaOH, we use this to work the Equillibirum moles
How? initial moles +/- Smallest Moles worked out
Equillibrium Moles:
NaOH = 0.0025-0.00025= 0
butanoic acid = 0.0125-0.0025= 0.01
Conjugate base (Salt) = 0 + 0.00025= 0.0025
If your wondering why i added for the salt and subtracted for the others is because, across the equilibrium sign, the sign changes and vice versa, but the smallest moles remains the same.
5) Calculating pH
-logKa + log(Conjugate base/Weak Acid)
We use the equilibrium moles for above as well as the given Ka which is 1.51x10-5
-log(1.51x10-5) + log(0.0025/0.01)
pH= 4.22
EDIT: I hope i got right lol
I got 4.22 and this is how i worked it.
1) First right out Equation
Butanoic Acid + NaOh <---> Conjugate Base (salt) + H20 (components)
2) For each Compenent we have a 1:1 ratio
3)We then work out Initial Moles of the given
So moles of Butanoic acid is 0.0125
And moles of NaOH is 0.0025
Current Moles of the Conjugate base (Salt), we assume is 0
4)We then use the smallest moles, in this case the moles for NaOH, we use this to work the Equillibirum moles
How? initial moles +/- Smallest Moles worked out
Equillibrium Moles:
NaOH = 0.0025-0.00025= 0
butanoic acid = 0.0125-0.0025= 0.01
Conjugate base (Salt) = 0 + 0.00025= 0.0025
If your wondering why i added for the salt and subtracted for the others is because, across the equilibrium sign, the sign changes and vice versa, but the smallest moles remains the same.
5) Calculating pH
-logKa + log(Conjugate base/Weak Acid)
We use the equilibrium moles for above as well as the given Ka which is 1.51x10-5
-log(1.51x10-5) + log(0.0025/0.01)
pH= 4.22
EDIT: I hope i got right lol
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niceguy95
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(Original post by langlitz)
We already knew the answer and how to do it lol (yes 4.22 is correct :L). We were just discussing the complexities of equilibria
We already knew the answer and how to do it lol (yes 4.22 is correct :L). We were just discussing the complexities of equilibria
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Borek
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#11
(Original post by Joynzer)
So if the strong base was in excess then there would be no buffer right?
So if the strong base was in excess then there would be no buffer right?
If however, we had a situation where the buffer was being formed from say butanoic acid and sodium bethanoate (and the sodium bethanoate was in excess) a buffer solution could still be formed?
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Joynzer
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#12
(Original post by Borek)
Yes, after neutralization that would be no acid left, you will have a mixture of conjugate base and excess strong base.
Yes, after mixing them you will have a solution containing both conjugate acid and conjugate base.
Yes, after neutralization that would be no acid left, you will have a mixture of conjugate base and excess strong base.
Yes, after mixing them you will have a solution containing both conjugate acid and conjugate base.
If the question asks about a buffer being formed from a weak acid and it's salt then either the acid or salt can be in excess. Also, the concentrations of the acid and conjugate base in the equilibrium mixture can be found by calculating the moles of the acid and conjugate base given (concentration x volume) and then by dividing by the total volume of the buffer.
If the question asks about a buffer being formed from a weak acid and it's reaction with a strong base (e.g. NaOH). The acid must be in excess for the buffer to form. The concentrations of the equilibrium mixture in this case would be found by working out the initial moles of acid (concentration x volume) and the base etc. We would then have to work out the final concentrations by taking in to account how much of the acid has reacted with the base (NaOH) to form the conjugate base.
Does that sound okay?
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langlitz
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#13
(Original post by Joynzer)
Okay so in conclusion.
If the question asks about a buffer being formed from a weak acid and it's salt then either the acid or salt can be in excess. Also, the concentrations of the acid and conjugate base in the equilibrium mixture can be found by calculating the moles of the acid and conjugate base given (concentration x volume) and then by dividing by the total volume of the buffer.
If the question asks about a buffer being formed from a weak acid and it's reaction with a strong base (e.g. NaOH). The acid must be in excess for the buffer to form. The concentrations of the equilibrium mixture in this case would be found by working out the initial moles of acid (concentration x volume) and the base etc. We would then have to work out the final concentrations by taking in to account how much of the acid has reacted with the base (NaOH) to form the conjugate base.
Does that sound okay?
Okay so in conclusion.
If the question asks about a buffer being formed from a weak acid and it's salt then either the acid or salt can be in excess. Also, the concentrations of the acid and conjugate base in the equilibrium mixture can be found by calculating the moles of the acid and conjugate base given (concentration x volume) and then by dividing by the total volume of the buffer.
If the question asks about a buffer being formed from a weak acid and it's reaction with a strong base (e.g. NaOH). The acid must be in excess for the buffer to form. The concentrations of the equilibrium mixture in this case would be found by working out the initial moles of acid (concentration x volume) and the base etc. We would then have to work out the final concentrations by taking in to account how much of the acid has reacted with the base (NaOH) to form the conjugate base.
Does that sound okay?
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Joynzer
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