Write up the recipe to make 100mL of a 200mM phosphate buffer, pH 7.0 by using:

Question

(a) only NaH2PO4 (MW 120g/mol) and NaOH (1.00M)

(b) only Na2HPO4 (MW 142g/mol) and HCl (1.00M)

(c) only Na2HPO4 and NaH2PO4

Please help me. I don't seem to understand how to even start......

Thanks

Answer 1

I didn't answer this last yesterday when you first posted because the post was screwed up. It looks ok now.

You want 100 mL of 200 mM phosphate buffer. That is 100 mL x 0.2 M = 20 millimols phosphate. So you know base(b) + acid (a) = 20. That is equation 1.

You obtain equation 2 from the HH equation of 7.00 = pKa2 + log (b/a) and solve that equation for b/a. That is equation 2. pKa2 is that for H3PO4.

Solve equation 1 and equation 2 simultaneously and determine b and a (they will be in millimols).

For c, which is the easiest of the 3, now that you know b and a (and those are in millimoles), calculate how much NaH2PO4 and Na2HPO4 that you must weigh out for the millimoles of each that you calculate. Obviously NaH2PO4 is the acid and Na2HPO4 is the base.

Then for part a and part b you go a slightly different route. I'll start you on part a and part b I'll leave to you. It's the same kind of thing.

......H2PO4^- + OH^- ==> HPO4^2- + H2O
I......20.......0..........0
add.............x............
C.....-x.......-x...........x
E......20-x.....0...........x
You are starting with 20 millimoles of the acid and x = millimols NaOH to be added.

Then substitute the E line into the HH equation like this
7.0 = pKa2 + log (x)/(20-x)
Solve for x = millimols NaOH. Now you know mmols NaOH and you know it is 1 M, solve for mL NaOH to add to that 20 mmols of the NaH2PO4.

For part B, you start with 20 millimols Na2HPO4 and add H^+ to make H2PO4^- but it's done exactly the same way but now you let x = millimols HCl that must be added.
I love these problems. By the way, the CORRECT way to do these is with concentrations but I've usee millimols. You can use concentration, mols, millimols BECAUSE (a) it is a ratio and (b) M = millimols/mL and since mL is the same for both acid and base and since they are in the denominator of base/acid, that means the denominator ALWAYS cancels so I just don't put the mL there in the first place.

Answer 2

Thank you so much!