pH on addition of NaOH

Apr 2014
2
0
Hi there,

I have a question which I am terribly stuck on!
I feel as though I have tried almost everything and cannot get any of the options given . I would appreciate any help that I can get!

I have attached the question below.

I used the Henderson-hasselbach equation pH=pka + log [A-]/[HA]

We know that ka = 6.31 x10 ^-6. I also made the assumption that 0.2mols is 0.2molL-1 (not sure if I can do that though)
Final pH = 5.2, so final [H3O+] = 6.31 x10^-6 so pka = final pH.

I managed to figure out the initial pH concentration which is Squareroot (Ka x [HA] = 1.12 x 10^-3 molL-1 then -log1.12 x 10-3 = 2.94947
Then I tried finding the difference in pH between final and initial = 5.20- 2.94947 = 2.2505 and assumed that the difference was the amount of NaOH added, but I'm completely lost. Not sure if I am doing this right at all, please help!
Thanks!
 

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Last edited:
May 2010
491
8
pH after NaOH addition

Ka =6.31x 10^-6 = (H^+) (A^-)/HA
pH =5.2 (H^+)= 10^-5.2
6.31x 10^-6/10^-5.2 = (A)/(0.2 -A)
1= (A)/(0.2-A)
A= 0.1 mol
0.2 -A = 0.1mol HA remaining