If 10mL of 1 M HCl is added to 1L of 0.01 M phosphate buffer, pH 7.2, what is the resulting pH?

Question

Also what are the conc of H2PO4^1- and HPO4^2- in the final solution?

HCl:
M = mol / L
1M = mol / 0.01L
= 0.01 mol

phosphate:
M = mol / L
0.01M = mol / 1L
= 0.01 mol

H2PO4 = 0.005mol
HPO4 = 0.005mol

so im lost here....

using HH equation
pH= 2.14 +log(0.005)/(0.005)
pH = 2.14

pH= 2.14 from question

thanks a lot

Answer 1

I think the pH of 2.14 is correct but this is really a complex question. As for the concentrations, H2PO4^- = 5 mmols/1010 mL = ?
HPO4^2- = 0
H3PO4 is 5 mmols/1010 mL but the problem doesn't ask for that.

I approached the problem this way. The HH equation with the pH of 7.20 is
7.20 = 7.20 + log (b/a) so b/a = 1
Then you know a + b = 0.01 x 1000 = 10 mmols and solving the two equations simultaneously, even though you already know it, a = b in mmols which means a = b = 5 mmols.. So
...............HPO4^2- + HCl ==> H2PO4^- + ...
I................5.................0...............5
add.............................10....................
C............-5.................-5...............+5
E............. 0..................5................10
It's important here to realize that there is enough HCl added to convert ALL of the HPO4^2- to H2PO4^- and have an excess of 5 mmols HCl remain which can then react with H2PO4^- to form H3PO4. You put those new values into a new HH equation, but this time with k1 and not k2. I hope this helps.