Back
to top
Calculate the ph f this solution
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
0.050moldm-3 koh titrated against 20cm3 0.020moldm-3 c6h5cooh untill neutralisation
the volume of 0.050moldm-3 koh needed in titration is Xcm3. Calc X. I got this which is 8cm3
The titration repeated a 2nd time using X/2 cm3 of 0.050moldm-3 koh. Calc the ph of this solution.
No idea how to do this
Any help appreciated
the volume of 0.050moldm-3 koh needed in titration is Xcm3. Calc X. I got this which is 8cm3
The titration repeated a 2nd time using X/2 cm3 of 0.050moldm-3 koh. Calc the ph of this solution.
No idea how to do this
Any help appreciated
0
reply
(Original post by User101010)
0.050moldm-3 koh titrated against 20cm3 0.020moldm-3 c6h5cooh untill neutralisation
the volume of 0.050moldm-3 koh needed in titration is Xcm3. Calc X. I got this which is 8cm3
The titration repeated a 2nd time using X/2 cm3 of 0.050moldm-3 koh. Calc the ph of this solution.
No idea how to do this
Any help appreciated
0.050moldm-3 koh titrated against 20cm3 0.020moldm-3 c6h5cooh untill neutralisation
the volume of 0.050moldm-3 koh needed in titration is Xcm3. Calc X. I got this which is 8cm3
The titration repeated a 2nd time using X/2 cm3 of 0.050moldm-3 koh. Calc the ph of this solution.
No idea how to do this
Any help appreciated
The answer is 4.20 btw but can't get to here
0
reply
Report
#3
(Original post by User101010)
The answer is 4.20 btw but can't get to here
The answer is 4.20 btw but can't get to here
To get the pH there must be a value for Ka or pKa for the acid.
In the second part when the titration is repeated with X/2 cm3 of KOH the acid will be half neutralised.
This means the concentrations of acid and its salt in the buffer are equal as half of the acid has neutralised and formed the salt plus water. The ratio of acid : salt is 1:1 so how ever many moles of acid are reacted the same number of moles of salt are formed.
Use the buffer equation
[H+] = Ka x ([acid]/[base])
because the concentrations of acid and base are the same ([acid]/[base])= 1
so [H+] = Ka this means pH = pKa.
0
reply
(Original post by Madasahatter)
Your answer for X = 8 cm3 is correct.
To get the pH there must be a value for Ka or pKa for the acid.
In the second part when the titration is repeated with X/2 cm3 of KOH the acid will be half neutralised.
This means the concentrations of acid and its salt in the buffer are equal as half of the acid has neutralised and formed the salt plus water. The ratio of acid : salt is 1:1 so how ever many moles of acid are reacted the same number of moles of salt are formed.
Use the buffer equation
[H+] = Ka x ([acid]/[base])
because the concentrations of acid and base are the same ([acid]/[base])= 1
so [H+] = Ka this means pH = pKa.
Your answer for X = 8 cm3 is correct.
To get the pH there must be a value for Ka or pKa for the acid.
In the second part when the titration is repeated with X/2 cm3 of KOH the acid will be half neutralised.
This means the concentrations of acid and its salt in the buffer are equal as half of the acid has neutralised and formed the salt plus water. The ratio of acid : salt is 1:1 so how ever many moles of acid are reacted the same number of moles of salt are formed.
Use the buffer equation
[H+] = Ka x ([acid]/[base])
because the concentrations of acid and base are the same ([acid]/[base])= 1
so [H+] = Ka this means pH = pKa.
So far in my syllabus I've only covered ph of acids and alkalis and ionisation of water is this why I'm unable to understand or should I know this from covering the topics I have
0
reply
Report
#5
(Original post by User101010)
So far in my syllabus I've only covered ph of acids and alkalis and ionisation of water is this why I'm unable to understand or should I know this from covering the topics I have
So far in my syllabus I've only covered ph of acids and alkalis and ionisation of water is this why I'm unable to understand or should I know this from covering the topics I have
If you can do pH of a weak acid you can do this though because
Ka = [H+][A-]/[HA] still holds true but for a weak acid [H+] = [A-] so when calculating pH we say ka = [H+]^2 / [HA] but in the example in the question [H+] does not equal [A-] because this is a partially neutralised acid. So you have to calculate [HA] and [A-] and then use
Ka = [H+][A-]/[HA] to calculate [H+] by rearranging it to
[H+] = Ka x [HA]/[A-]
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
