dxnixl
Badges: 18
Rep:
?
#1
Report Thread starter 1 month ago
#1
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
Name:  phhelp.PNG
Views: 8
Size:  99.3 KB

Thanks
0
reply
*****deadness
Badges: 19
Rep:
?
#2
Report 1 month ago
#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
1
reply
dxnixl
Badges: 18
Rep:
?
#3
Report Thread starter 1 month ago
#3
(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
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?
0
reply
*****deadness
Badges: 19
Rep:
?
#4
Report 1 month ago
#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?
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
0
reply
dxnixl
Badges: 18
Rep:
?
#5
Report Thread starter 1 month ago
#5
(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
oopsy silly me yeah its HNO3 Thankyou for this!!
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Have you experienced financial difficulties as a student due to Covid-19?

Yes, I have really struggled financially (20)
13.42%
I have experienced some financial difficulties (41)
27.52%
I haven't experienced any financial difficulties and things have stayed the same (61)
40.94%
I have had better financial opportunities as a result of the pandemic (23)
15.44%
I've had another experience (let us know in the thread!) (4)
2.68%

Watched Threads

View All