[ccrec]

Hi.  Having some trouble with a chem problem.  The question is:  what weight of solid sodium hydroxide must be added to 600ml of 0.1M NaH2(PO)4 to give a sodium phosphate buffer w/ a pH of 7.4 (assume negligible volume change).  I have been given that the correct answer is 1.918g.

So, I worked out that my 600ml of 0.1M soln has 7.2g NaH2(PO)4.  I assume I have to use Henderson-Hasselbach, but when I set it up as 7.4=12+log[NaOH]/[NaH2(PO)4] I don't get the right answer.  I am setting up this way b/c 12 is the third pKa of NaH2(PO)4.  But I don't get the right answer when I use the 2nd pKa either (6..  I assume I am doing something wrong with the setup of my equation, but can't figure out what...please *delete me*!  thanks!

[Borek]

when I set it up as 7.4=12+log[NaOH]/[NaH2(PO)4] I don't get the right answer

NaOH and NaH₂(PO)₄ are not a pair of acid and conjugate base.

http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch

also check

http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-buffer-q2

Although it deals with other buffer and HCl (not NaOH) addition, it is about the same problem.

Second pKa is the one you have to use.

[oldddog]

You should write out the reaction equation to help with the solution to this problem; (you can get rid of the spectator ions if you like.

NaOH       +    NaH₂PO₄   ---->   Na₂HPO₄    +    NaH₂PO₄     +    H₂O
  ?                 in excess        conjugate base    conjugate acid

You then use the Henderson-Hasselbach equation to determine the concentration of HPO42- required for this buffer solution, hence the amount of NaOH required to form it.

[FeLiXe]

You want to have an H₂PO₄^-/HPO₄^2- buffer

the latter you get when you add NaOH:
OH^- + H₂PO₄^- <-> HPO₄^2- + H₂O

The little trick with that kind of question is now that as you add NaOH, the NaOH goes away and you get the same amount of HPO₄^2- and at the same time you get less H₂PO₄^-

The table would look like this:

	OH- +	H2PO4- <->	HPO42- +	H2O
before reaction:	x	a	0	-
after reaction:	0	a-x	x	-

x is the amount of NaOH you added
a is the amount of H₂PO₄^- that was in the solution at first

[sdekivit]

hmmmm my answer is 1,46 g  

there is the following equilibrium:

H2PO4(-) + H2O <--> HPO4(2-) + H3O(+)

Ka = [HPO4(2-)][H3O(+)]/[H2PO4(-)] = 6,2 x 10^-8

--> [HPO4(2-)]/[H2PO4(-)] = 6,2 x 10^-8 / 10^-7,4 = 1,55737

we started with 60 mmol H2PO4(-) after reaction there is left 60 - x mmol H2PO4(-) and x mmol HPO4(2-) is formed

--> x / (60 - x) = 1,55737 --> x = 36,53849 mmol H2PO4(-) had reacted with OH(-) thus we need 36,53849 * M(NaOH)/1000 = 1,46 g NaOH

[Borek]

hmmmm my answer is 1,46 g  

1.47g here - so the answer given to Claudia must to be wrong.

The difference (1.46/1.47) is neglectable and is probably due to used pKa2 value and due to the fact that I am using precise molar masses (not that I am so precise, they are just built into my programs).