Buffer Help?

Question

What volume (to the nearest 0.1 mL) of 4.30-M HCl must be added to 0.350 L of 0.200-M K2HPO4 to prepare a pH = 7.80 buffer?

I keep getting 13mL by making a rxn table and finding x. Then converting from L-Ml I am doing something wrong.

PH=Pka+log(b/a)
H3O+HPO4<->H30+H2PO4
x........(.07)........n/a........0
-x........-x............n/a........+x
0........(.07-x).....----........x

I found x to be .0557mols

.0557mols=4.30M * (#L)
(#L)=.01295L *1000mL=12.95mL

What am I doing wrong?

Answer 1

Yes, you are doing some thing wrong.
1) You should note that 0.350L*(0.200mol/L) = 0.07 mol, which is amount of K2HPO4. But doing this kind of calculation, you should use CONCENTRATION. If you still do not understand, think in extreme cases, when this 0.07 mole K2HPO4 is in an occion.
Please follow me.
Let X L be the volume of 4.30-M HCl to be added.
Notice that the concentration of chemicals would be changed by change the volume of the solution.
The new K2HPO4 concentration before reaction: 0.07/(X+0.35).
The new HCl concentration before reaction: 4.30*X/(X+0.35).
After reaction, the KH2PO4 concentration is roughly: 4.30*X/(X+0.35).
and the left K2HPO4 concentration is: (0.07-4.30*X)/(X+0.35).
H2O + H2PO4(-) <==> H3O(+) + HPO4(2-), Ka
Now you have (notice HPO4(2-) is "b" and H2PO4(-) is "a"):
pH=pKa + log(b/a)
7.80 = -log(6.2x10^-8) + log{(0.07-4.30*X)/(4.30*X)}
(0.07-4.30*X)/(4.30*X) = 10^(0.5924) = 3.912
0.07 - 4.30*X = 16.82*X
X = 0.00331 (L) = 3.3 ml
That is the answer.