[Ann1234]

Hello =) I am studying acid-base titrations and I found this problem on chemwiki that is driving me crazy  . I solved and I got a different answer so if someone can check my answer and their answer and let me know if I am right or wrong...it will be great ...

What is the pOH when 5.0 L of a 0.45 M solution of sulfuric acid (H2SO4) is titrated with 2.3 L of a 1.2 M lithium hydroxide (LiOH) solution?

They got POH= 13.4, I got POH= 12.43.

This is what THEY did:

1) they calculated the moles of protons H+ from the total moles of sulfuric acid:

2) 0.45 MX5LX2molH/1mol H2SO4 = 4.5 moles

3) and then the moles of LiOH = 1.2M X 2.3L= 2.8 moles

4) then they calculated moles excess H+ = 4.5 - 2.8 = 1.7 MOLES H+

5) Then using total volume = 5 L + 2.3 l = 7.3 L they got the concentration of H+ (1.7/7.3 = 0.23M)

6) PH = -log (0.23) = 0.64 => POH = 14-0.64 = 13.4

I solved in a completely different way. I think they are wrong in considering that all the sulfuric acid will be dissociated in H+ since the amount of LIOH was not enough for neutralizing all the acid.(some acid was left in excess)

So this is what I did: I assumed that OH- will neutralize the H2SO4 first:

moles H2SO4 (initial) = 5X0.45 = 2.25 moles
moles LiOH (initial) = 2.3 X 1.2 = 2.76 moles

H2SO4 first dissociation is considered strong:

1) H2SO4 + OH- ----> HS04- + H20
I     2.25      2.76      ----      ----
C   -2.25    -2.25      2.25      ----
E    -----    0.51       2.25     ----

So after the first neutralization, we have an excess of OH- and all H2SO4 was converted to HSO4-. Now, another ice table with HSO4- neutralization:

2) HSO4- + OH- = SO4-2 + H20
I   2.25      0.51    -----     -----
C  -0.51    -0.51   +0.51    -----
E  1.74     -----    0.51     -----

So finally, after all the moles of LiOH reacted, we have 1.74 moles HSO4- in solution and 0.51 moles SO4-2. HS04- is a weak acid, so to calculate PH we need to consider its ka = 0.012. Also we need the concentration of HS04 and SO4-2:

[HSO4-]= 1.74/7.3L= 0.2384 M
[SO4-2] = 0.51/7.3L = 0.06986

   HSO4-      =    H+    +    SO4-2
I  0.2384            ---      0.06986
C  -X                 +X         +X
E (0.2384 - X)     (X)    (0.06986 + X)

Ka = 0.012 = X. ( 0.06986+ X)/(0.2384 - X)

I used the successive approximation method and I got x= 0.027

x = [H+] = 0.027

PH = -log (0.027) = 1.57 => POH = 12.43


I know is a lot of reading so thank you so much for taking the time to read my problem and hopefully you guys can give me some feedback, I would really like to know if I did this problem right.

Thank you =)

[Borek]

You are right.

[Ann1234]

You are right.

oh awesome... I also continued one more time with the successive aproximation and I got x= 0.026 which gives me a PH=1.58  

Thanks for the confirmation-that ph software rocks!  

[Borek]

This is BATE. Note that the screen shot is from the beta version that is not available yet. Old version has no indicator picture, but all calculations are done the same way.

[Ann1234]

This is BATE. Note that the screen shot is from the beta version that is not available yet. Old version has no indicator picture, but all calculations are done the same way.

okay =) I was browsing your website and I have a question...this PH example:

http://www.chembuddy.com/?left=BATE&right=example_of_pH_calculation

pH calculation of 0.01M (NH4)2HPO4 solution, that is from a weak acid and the amphoteric acid HPO4-, you got PH=8.06 and is treated as a titration H3PO4- vs NH3.

I tried using the formula for amphiprotic salts simplified

http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified

and I got 8.22, probably higher than the real one since I neglected the HPO4-2 dissociation (so I just considered NH4+ as an acid and HPO4- as a base).

Is there a possibility of getting this PH using a formula?

thank you 

[Borek]

I tried using the formula for amphiprotic salts simplified

http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified

and I got 8.22, probably higher than the real one since I neglected the HPO4-2 dissociation (so I just considered NH4+ as an acid and HPO4- as a base).

Is there a possibility of getting this PH using a formula?

No - it is a simplified formula, so it gives only approximate results. Quite good taking into account all equilibria present.

[Ann1234]

oh okay...thanks that's good to know, I was burning my brain cells trying to come up with that answer!