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Hi,
I don't get how to do this question and don't yet know how to read off titration curve graphs? Could someone help? I'll attach it below
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Thanks
I don't get how to do this question and don't yet know how to read off titration curve graphs? Could someone help? I'll attach it below
Thanks
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#2
For part b use the end point, which is where the vertical region is. For this graph this is 25cm3. We know the mole ratio of HNO3:NaOH, so find the concentration that way.
For part c this isn't actually a buffer because although there is sodium nitrate in the solution, we also have NaOH which is a strong alkali and it doesn't not form a buffer. I do edexcel and we ignore the salt, but I'm not sure if you approximate it the same way with OCR. If they do, think of what is left in the solution when 50cm3 is added and calculate the concentration ignoring the salt
For part c this isn't actually a buffer because although there is sodium nitrate in the solution, we also have NaOH which is a strong alkali and it doesn't not form a buffer. I do edexcel and we ignore the salt, but I'm not sure if you approximate it the same way with OCR. If they do, think of what is left in the solution when 50cm3 is added and calculate the concentration ignoring the salt
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(Original post by *****deadness)
For part b use the end point, which is where the vertical region is. For this graph this is 25cm3. We know the mole ratio of HNO3:NaOH, so find the concentration that way.
For part c this isn't actually a buffer because although there is sodium nitrate in the solution, we also have NaOH which is a strong alkali and it doesn't not form a buffer. I do edexcel and we ignore the salt, but I'm not sure if you approximate it the same way with OCR. If they do, think of what is left in the solution when 50cm3 is added and calculate the concentration ignoring the salt
For part b use the end point, which is where the vertical region is. For this graph this is 25cm3. We know the mole ratio of HNO3:NaOH, so find the concentration that way.
For part c this isn't actually a buffer because although there is sodium nitrate in the solution, we also have NaOH which is a strong alkali and it doesn't not form a buffer. I do edexcel and we ignore the salt, but I'm not sure if you approximate it the same way with OCR. If they do, think of what is left in the solution when 50cm3 is added and calculate the concentration ignoring the salt
b) Molar Mass = 0.1 x 0.02 = 0.002 moles
25cm3 = 0.025dm3
Therefore concentration = 0.08moldm3
c) I think (as it’s 4 marks) we do something different here:
50cm3 of 0.08moldm3 NaOH added to 20cm3 of 0.1moldm3 HCl
1) Mol NaOH = (50/1000)x(0.08) = 0.004 moles
2) Mol HCl = (20/1000)x(0.1) = 0.002 moles
3) Extra NaOH = 0.004-0.002 = 0.002 moles
4) Concentration = (0.002/0.07) moldm3
5) Ph = -log(10^-14/(0.002/0.07)) = 12.46
PH = 12.46
is that right?
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#4
(Original post by dxnixl)
Ah okay soo
b) Molar Mass = 0.1 x 0.02 = 0.002 moles
25cm3 = 0.025dm3
Therefore concentration = 0.08moldm3
c) I think (as it’s 4 marks) we do something different here:
50cm3 of 0.08moldm3 NaOH added to 20cm3 of 0.1moldm3 HCl
1) Mol NaOH = (50/1000)x(0.08) = 0.004 moles
2) Mol HCl = (20/1000)x(0.1) = 0.002 moles
3) Extra NaOH = 0.004-0.002 = 0.002 moles
4) Concentration = (0.002/0.07) moldm3
5) Ph = -log(10^-14/(0.002/0.07)) = 12.46
PH = 12.46
is that right?
Ah okay soo
b) Molar Mass = 0.1 x 0.02 = 0.002 moles
25cm3 = 0.025dm3
Therefore concentration = 0.08moldm3
c) I think (as it’s 4 marks) we do something different here:
50cm3 of 0.08moldm3 NaOH added to 20cm3 of 0.1moldm3 HCl
1) Mol NaOH = (50/1000)x(0.08) = 0.004 moles
2) Mol HCl = (20/1000)x(0.1) = 0.002 moles
3) Extra NaOH = 0.004-0.002 = 0.002 moles
4) Concentration = (0.002/0.07) moldm3
5) Ph = -log(10^-14/(0.002/0.07)) = 12.46
PH = 12.46
is that right?
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(Original post by *****deadness)
Other than the fact you've used HCl instead of HNO3, I see no problem with it, although you could've done (50-25)/1000*0.08 to work out the extra NaOH directly and save a step. But what you did is perfectly fine
Other than the fact you've used HCl instead of HNO3, I see no problem with it, although you could've done (50-25)/1000*0.08 to work out the extra NaOH directly and save a step. But what you did is perfectly fine
Thankyou for this!!
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