How would you prepare a 1.0 L buffer solution of pH 6.0 by mixing 0.50 M Na3PO4 and 0.40 M HCl?
pH = pKa2 + log (base/acid)
6.0 = 7.20 + log (b/a)
log b/a = 6.0-7.20 = -1.20
b/a = 0.063 and
b = 0.063*a This is equation 1.
Let's make the buffer solution = 1 M. equation 2 is
a + b = 1 and solve these two equations simultaneously
a + 0.063a = 1
1.063a = 1
a = 1/1.063 = 0.941 M
b = 1- 0.941 = 0.059 M
Since this is 1 L of solution, that will be 0.941 mols acid (HPO4^-) and 0.059 mols base (PO4^3-). Therefore, what do we start with to have these mols at the end. We start with 1 mols Na3PO4 and add 0.059 mols HCl and make the volume up to 1000 mL.
Now calculate how many mL of the 0.50 M Na3PO4 you must add to add 1 mol Na3PO4 and how many mL of the 0.40 M HCl you must add to add 0.941 mols. Post your work if you get stuck.
You may want to check the results:
pH = 7.20 + log b/a
pH = 7.20 + log (0.059/0.941) = 5.997 = 6.0
Is it obvious to you that you must start with 1 mol Na3PO4 and add 0.941 mol HCl. Do it this way.
...............PO4^3- + H^+ ==> HPO4^-
I.................x............0................0
add...........................y.........................
C..............-y..............-y................+y
E.............x-y...............0.................y
You want to end up with acid (HPO4^-) to be 0.941 so y = 0.941.
You want to end up with x-y (PO4^-) to be 0.059. So if
x-y = 0.059 and y is 0.941, then x, the starting amount of Na3PO4 must be x = 0.059 + 0.941 = 1.
Good luck.