Question

For a buffer of H2PO4- and HPO42- with concentrations of .4M and .6M respectively with a volume of 250mL.

1) What is the initial pH?
2) How much of what can be added for the pH to be equal to the pKa?
3) If .01M HCl is added, how much is required to reach the equivalence point?
4) If 5mL of this HCl solution is added, what is the pH?

Answer 1

For the given buffer,

1) initial pH = pKa + log([HPO4^2-]/[H2PO4-])

Feeding the given values,

pH = 7.20 + log(0.6/0.4) = 7.38

2) For pH = pKa we have to make [H2PO4-] = [HPO4^2-]

So, adding 0.025 mols of HCl to the solution would make equal concentrations of both.

and thus pH = pKa

3) At equivalence point

all of HPO42- is converted to H2PO4^-

moles of HPO42- present = 0.6 M x 0.25 L = 0.15 mol

So we have to add 0.15 mol of HCl to the solution

Volume of 0.01 M HCl to be added = 0.15 mol/0.01 M = 15 ml

4) If 5 ml of HCl is added

final [H2PO4-] = (0.4 x 0.25 + 0.01 x 0.005)/0.255 = 0.392 M

final [HPO4^2-] = (0.6 x 0.25 - 0.01 x 0.005)/0.255 = 0.588 M

pH = 7.20 + log(0.588/0.392) = 7.37