Alevel chemistry buffers Watch

sohaail23
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In some buffer questions you have to look at which is in excess then find the excess moles and convert it into concentration and then sub the values into the ka formula,

The acid dissociation constant, Ka, for the weak acid HY has the value
1.35 × 10–5 mol dm–3 at 25 °C.
A buffer solution was prepared by dissolving 0.0236 mol of the salt NaY in
50.0 cm3
of a 0.428 mol dm–3 solution of the weak acid HY
(i) Calculate the pH of this buffer solution.

For this question they didnt look at which was in excess, could anyone explain why?
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jAlAl12362
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I thought you would need excess weak acid, but if you work out the moles of the acid used you get 0.0214 which is < 0.0236 of the salt.
How is that even possible to make a buffer?
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Trapmoneybenny
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(Original post by sohaail23)
In some buffer questions you have to look at which is in excess then find the excess moles and convert it into concentration and then sub the values into the ka formula,

The acid dissociation constant, Ka, for the weak acid HY has the value
1.35 × 10–5 mol dm–3 at 25 °C.
A buffer solution was prepared by dissolving 0.0236 mol of the salt NaY in
50.0 cm3
of a 0.428 mol dm–3 solution of the weak acid HY
(i) Calculate the pH of this buffer solution.

For this question they didnt look at which was in excess, could anyone explain why?
The questions in which they present a scenario in which one of the reagents of the buffer is in excess is different from the one which you just posted.

This is a buffer containing a weak acid and it's salt ONLY. It's an acidic buffer, which explains why the pH is 4.91 (2dp) In this scenario the component in excess is irrelevant because the buffer is formed AFTER addition of the soluble salt.

Now if there was a buffer solution containing the components above and something like NaOH was added you would then need to calculate which one was in excess to find out the new pH

hope that clarifies
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sohaail23
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(Original post by jAlAl12362)
I thought you would need excess weak acid, but if you work out the moles of the acid used you get 0.0214 which is < 0.0236 of the salt.
How is that even possible to make a buffer?
when i tried i used excess NaY and found the moles in excess and used it in my KA but in the markscheme they never did this
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sohaail23
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(Original post by Oruese)
The questions in which they present a scenario in which one of the reagents of the buffer is in excess is different from the one which you just posted.

This is a buffer containing a weak acid and it's salt ONLY. It's an acidic buffer, which explains why the pH is 4.91 (2dp) In this scenario the component in excess is irrelevant because the buffer is formed AFTER addition of the soluble salt.

Now if there was a buffer solution containing the components above and something like NaOH was added you would then need to calculate which one was in excess to find out the new pH


hope that clarifies

So this question the buffer is there and these are the concentrations oin the buffer and i just need to find the ph
in others i need to find the the concentrations in the buffer?
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jAlAl12362
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Ok, made a mistake.

You use this formula (Ka)*(acid)/ (salt)

You got the concentration of acid given which is 0.428.
All you need to do is work out the concentration of salt which is 0.0236/0.05 Remember c*v= moles just rearrange it.

Add that to your formula to work out H+ concentration which I got was 1.2242*10-5

Do -log(H+).
I got pH of 4.912
Is that right?
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sohaail23
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(Original post by Oruese)
The questions in which they present a scenario in which one of the reagents of the buffer is in excess is different from the one which you just posted.

This is a buffer containing a weak acid and it's salt ONLY. It's an acidic buffer, which explains why the pH is 4.91 (2dp) In this scenario the component in excess is irrelevant because the buffer is formed AFTER addition of the soluble salt.

Now if there was a buffer solution containing the components above and something like NaOH was added you would then need to calculate which one was in excess to find out the new pH

hope that clarifies
For this question would i need to look at whats in excess? could you explain whether i do or not using your example from before please

Calculate the pH of the buffer solution formed when 10.00 cm3 of 0.100 mol dm–3 potassium hydroxide are added to 25.00 cm3 of 0.410 mol dm–3 ethanoic acid.
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sohaail23
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(Original post by jAlAl12362)
Ok, made a mistake.

You use this formula (Ka)*(acid)/ (salt)

You got the concentration of acid given which is 0.428.
All you need to do is work out the concentration of salt which is 0.0236/0.05 Remember c*v= moles just rearrange it.

Add that to your formula to work out H+ concentration which I got was 1.2242*10-5

Do -log(H+).
I got pH of 4.912
Is that right?
Answer was 3.79
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sohaail23
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(Original post by sohaail23)
For this question would i need to look at whats in excess? could you explain whether i do or not using your example from before please

Calculate the pH of the buffer solution formed when 10.00 cm3 of 0.100 mol dm–3 potassium hydroxide are added to 25.00 cm3 of 0.410 mol dm–3 ethanoic acid.
(Original post by jAlAl12362)
Ok, made a mistake.

You use this formula (Ka)*(acid)/ (salt)

You got the concentration of acid given which is 0.428.
All you need to do is work out the concentration of salt which is 0.0236/0.05 Remember c*v= moles just rearrange it.

Add that to your formula to work out H+ concentration which I got was 1.2242*10-5

Do -log(H+).
I got pH of 4.912
Is that right?
http://pmt.physicsandmathstutor.com/...n%203%20MS.pdf

Q1 D
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Trapmoneybenny
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(Original post by sohaail23)
For this question would i need to look at whats in excess? could you explain whether i do or not using your example from before please

Calculate the pH of the buffer solution formed when 10.00 cm3 of 0.100 mol dm–3 potassium hydroxide are added to 25.00 cm3 of 0.410 mol dm–3 ethanoic acid.
i need the Ka for ethanoic acid though, what's the value they gave ya?
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jAlAl12362
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Markscheme says im right, which question is this?
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