frodo00
Badges: 8
Rep:
?
#1
Report Thread starter 1 year ago
#1
the answer is 4.20, I got 13.70


the question is find the pH of 0.100moldm-3 benezencarboxylic acid (Ka=6.31X10-5) when it has half been neutralised by NaOH


Any help would be appreciated. Thanks x
0
reply
BobbJo
Badges: 12
Rep:
?
#2
Report 1 year ago
#2
Only half has been neutralised... The pH must be acidic. It cannot be > 7 as you got

Half of benzoic acid is neutralised

so [C6H5COOH] = 0.05 M

when benzoic acid reacts with NaOH, the sodium salt is formed. hence

[C6H5COONa] = 0.05 M

[C6H5COOH + NaOH -> C6H5COONa + H2O]

replace into Ka expression

pH = 4.2
2
reply
charco
Badges: 17
Rep:
?
#3
Report 1 year ago
#3
(Original post by frodo00)
the answer is 4.20, I got 13.70


the question is find the pH of 0.100moldm-3 benezencarboxylic acid (Ka=6.31X10-5) when it has half been neutralised by NaOH


Any help would be appreciated. Thanks x
... and the trick is ...

at the half equivalence point pH = pKa
1
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

Are you travelling in the Uni student travel window (3-9 Dec) to go home for Christmas?

Yes (107)
28.23%
No - I have already returned home (50)
13.19%
No - I plan on travelling outside these dates (71)
18.73%
No - I'm staying at my term time address over Christmas (39)
10.29%
No - I live at home during term anyway (112)
29.55%

Watched Threads

View All