Question

This question pertaining to buffer systems!

How would you prepare 10.0mL of 0.0100M phosphate buffer, pH = 7.4
0, from stock solution of 0.100M KH2PO4 and 0.250M K2HPO4, pKa of KH2PO4 is 7.20.

Please help me in figuring out how I even start this problem...as well as attempting to finish it...

Answer 1

I like to do these in millimoles. That way I don't get bugged by the zeros in front.
pH = pK2 + log (base)/(acid)
mmols you want = 10 x 0.01 = 0.1.
So equation 1 is a + b = 0.1

Equation 2 is
7.4 = 7.2 + log (b)/(a)
Solve for b/a and rearrange to
b = ?*a (whatever that ? number is). I think that should be close to 5 = b/a or a = 5b.
Solve the two equations simultaneously for a and b.
When you know how many millimols a and mmols b you need, calculate mL you need of each of the stock solutions to give you that man mmols of each. Piece a cake. Post your work if you get stuck.

Answer 2

I was able to do Equation 2 and got 1.58 for b/a ...

Answer 3

pH = pK2 + log b/a
7.4 = 7.2 + log b/a
log b/a = 1.58. You're right, I typed in the wrong numbers in my calculator. I typed in 7.2= 7.4 + log b/a. It helps to use the right numbers doesn't it.
OK. b/a = 1.58 or b = 1.58a
then a + b = 0.1
a + 1.58a = 0.1 and I suspect you can take from there to find a and b (and those will be in mmoles).
From there you will know mmols a, mmols b, and the stock solutions are
M = mmols/mL. You will know mmols a and M of that stock, substitute and solve for mL.

Answer 4

I was able to solve it! Thank you again Dr.Bob!
This is the 2nd time you've saved me this semester!
I greatly appreciate it!
:D

Answer 5

That's great. Glad to help. Remember to look to see the METHOD so you will remember it.