Ratio in buffer calculation please help me

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For this question, the ratio is already given in part b. I don't really get how the conc of the two carbonate ions are worked out. Because from my calculation I got for the total mole of acid. Surely the mole ratio of acid: carbonate ions should be 3:1 right? So how can the mole of the carbonate ions can be worked out by doing 4.4*10^-3 divided by /2.958? Why can't you find the mole of the carbonate ions by dividing by 3?

Thanks

(edited 7 years ago)

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Reply 1

coconut64

OP

16

Attachment not found

here is my working thanks a lot

(edited 7 years ago)

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Original post by coconut64

Attachment not found

For this question, the ratio is already given in part b. I don't really get how the conc of the two carbonate ions are worked out. Because from my calculation I got for the total mole of acid. Surely the mole ratio of acid: carbonate ions should be 3:1 right? So how can the mole of the carbonate ions can be worked out by doing 4.4*10^-3 divided by /2.958? Why can't you find the mole of the carbonate ions by dividing by 3?
Thanks

You are correct that the moles of acid used is 4.4 x 10^-3

Now you can see that 1 mol acid is equivalent to 3 mol of carbonate + hydrogencarbonate

Hence moles of carbonate + hydrogen carbonate in 10ml = 1.47 x 10^-3

You are now told that the pH of the lake is 10.3, which tells you the [H+] = 5.01 x 10^-11

and as

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

etc ...

Reply 3

Pigster

20

Original post by charco

Now you can see that 1 mol acid is equivalent to 3 mol of carbonate + hydrogencarbonate

Am I misunderstanding?

Original post by Pigster

Am I misunderstanding?

No, I am just expressing it the wrong way round ...

Moles of acid divided by 3 = total moles of carbonate plus hydrogen carbonate

Reply 5

coconut64

OP

16

Original post by charco

No, I am just expressing it the wrong way round ...

Moles of acid divided by 3 = total moles of carbonate plus hydrogen carbonate

Yes, that's what I did. I used the ratio to find the volume of each of the carbonate ions. So after that I used the volume and the mole to work out the conc of the carbonate ions. But why is my final answers wrong?

Thanks

Reply 6

coconut64

OP

16

Original post by charco

You are correct that the moles of acid used is 4.4 x 10^-3

Now you can see that 1 mol acid is equivalent to 3 mol of carbonate + hydrogencarbonate

Hence moles of carbonate + hydrogen carbonate in 10ml = 1.47 x 10^-3

You are now told that the pH of the lake is 10.3, which tells you the [H+] = 5.01 x 10^-11

and as

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

etc ...

This may be a bit freaky but I found one of your thread on the exact question but you have used two different methods here. https://www.thestudentroom.co.uk/showthread.php?t=2471172

In the older thread, you worked out the mole of the carbonate ions by working the ratio then work out the mole of the carbonate ions.
Hence moles of acid neutralised in terms of x = (2 + 0.958)x = 2.958x mol

Hence 2.958x = 0.0044 mol

x = 0.00149 mol

Therefore moles of carbonate in 10ml = 0.00149 mol
Therefore moles of bicarbonate in 10ml = 0.00143 mol

Can you explain why my method doesn't work? You can see it if you click on the attachment file.Thanks a lot.

(edited 7 years ago)

Original post by coconut64

This may be a bit freaky but I found one of your thread on the exact question but you have used two different methods here. https://www.thestudentroom.co.uk/showthread.php?t=2471172

In the older thread, you worked out the mole of the carbonate ions by working the ratio then work out the mole of the carbonate ions.
Hence moles of acid neutralised in terms of x = (2 + 0.958)x = 2.958x mol

Hence 2.958x = 0.0044 mol

x = 0.00149 mol

Therefore moles of carbonate in 10ml = 0.00149 mol
Therefore moles of bicarbonate in 10ml = 0.00143 mol

Can you explain why my method doesn't work? You can see it if you click on the attachment file.Thanks a lot.

There is often more than one way to skin a cat (apologies to vegans).

I showed this method in response to the questions asked about the method used here.

If you are getting the wrong answer then you must be making an error en-route.

Busy at the moment, but I'll look at it later.

Reply 8

coconut64

OP

16

Original post by charco

There is often more than one way to skin a cat (apologies to vegans).

I showed this method in response to the questions asked about the method used here.

If you are getting the wrong answer then you must be making an error en-route.

Busy at the moment, but I'll look at it later.

Yes I do realise my answer is wrong, but I dont know where I have gone wrong.

Thanks

Original post by coconut64

Yes I do realise my answer is wrong, but I dont know where I have gone wrong.

Thanks

Part (b)

[carbonate]/[hydrogencarbonate] = 0.958

pH = 10.3

Therefore [H+] = 5.01 x 10^-11

for the equation:

HCO₃- <==> CO₃^2- + H+

k = [H+][carbonate]/[hydrogencarbonate] = 0.958 x 5.01 x 10^-11 = 4.80 x 10^-11

Original post by coconut64

Yes I do realise my answer is wrong, but I dont know where I have gone wrong.

Thanks

Part (c)

Total moles of acid = 4.4 x 10^-3 (in 10ml)

Hence total moles of carbonate + hydrogen carbonate = (4.4 x 10^-3)/3 = 1.47 x 10^-3

Let mol carbonate = x
Let mol hydrogen carbonate = y

x + y = 1.47 x 10^-3
x/y = 0.958

x = 0.958y

0.958y + y = 1.47 x 10^-3

y = (1.47 x 10[sup]-3[/sup)/1.958 = 7.49 x 10^-4
x = 7.18 x 10^-4

both of these are in 10ml

Hence concentration in lake:

[carbonate] = 7.18 x 10^-2 mol dm^-3

[hydrogencarbonate] = 7.49 x 10^-2 mol dm^-3

(edited 7 years ago)

Reply 11

coconut64

OP

16

[QUOTE="charco;68906720"]Part (c)

Total moles of acid = 4.4 x 10^-3 (in 10ml)

Hence total moles of carbonate + hydrogen carbonate = (4.4 x 10^-3)/3 = 1.47 x 10^-3

Let mol carbonate = x
Let mol hydrogen carbonate = y

x + y = 1.47 x 10^-3
x/y = 0.958

x = 0.958y

0.958y + y = 1.47 x 10^-3

y = (1.47 x 10^{-3[/sup)/1.958 = 7.49 x 10^-4
x = 7.18 x 10^-4

both of these are in 10ml

Hence concentration in lake:

[carbonate] = 7.18 x 10^-2 mol dm^-3

[hydrogencarbonate] = 7.49 x 10^-2 mol dm^-3}

Really appreciate you taking your time to help. However I noticed your answer this time is different to the correct answer...

Therefore moles of carbonate in 10ml = 0.00149 mol
Therefore moles of bicarbonate in 10ml = 0.00143 mol
This is what you said in the older thread... But thanks anyway.

Original post by coconut64

Really appreciate you taking your time to help. However I noticed your answer this time is different to the correct answer...

Therefore moles of carbonate in 10ml = 0.00149 mol
Therefore moles of bicarbonate in 10ml = 0.00143 mol
This is what you said in the older thread... But thanks anyway.

After due consideration I see that my early assumption:

Hence total moles of carbonate + hydrogen carbonate = (4.4 x 10^-3)/3 = 1.47 x 10^-3

Is not valid:

If you let moles of carbonate = x and mol bicarbonate = y

Mol acid needed to neutralise x mol carbonate = 2x
Mol acid needed to neutralise y mol bicarbonate = y

Hence: 2x + y = 4.4 x 10^-3 mol

and x/y = 0.958, hence x = 0.958y

Substitute for x in the first equation

2(0.958y) + y = 4.4 x 10^-3

2.916y = 4.4 x 10^-3

y = 1.51 x 10^-3

x = 1.45 x 10^-3

[carbonate] = 0.145 mol dm^-3
[bicarbonate] = 0.151 mol dm^-3

(edited 7 years ago)

Reply 13

coconut64

OP

16

Original post by charco

What is the "correct" answer?

Therefore moles of carbonate in 10ml = 0.00149 mol
Therefore moles of bicarbonate in 10ml = 0.00143 mol

Should be the same answer as the one you posted in the older thread. Thanks

Original post by coconut64

Therefore moles of carbonate in 10ml = 0.00149 mol
Therefore moles of bicarbonate in 10ml = 0.00143 mol

Should be the same answer as the one you posted in the older thread. Thanks

corrected, slight differences are due to rounding up.

Ratio in buffer calculation please help me

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