# Calculate the ph f this solution

#1
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
#2
(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

The answer is 4.20 btw but can't get to here
0
7 years ago
#3
(Original post by User101010)
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
#4

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
7 years ago
#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
Presumably you will be doing buffers shortly if you have done pH of strong acids/alkalis and weak acids as that is what would normally come next.
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
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