Question

Using a 0.25 M phosphate buffer with a pH of 7.6, you add 0.75 mL of 0.45 M NaOH to 59 mL of the buffer. What is the new pH of the solution?

Answer 1

The buffer we are dealing with is composed mainly by the following species:

and it has a pKa = 7.21

To calculate the relation between HPO₄^-2 and H₂PO₄^- , we can use the hendersson hasselbach equation:

Where 7.21 is the pKa, [A-] = HPO₄^-2 and [AH]= H₂PO₄^-

so, pH = 7.6 = 7.21 + Log ([HPO₄^-2] / [H₂PO₄^- ] )

[HPO₄^-2] / [H₂PO₄^- ] = 2.45

If you add 0.75 mL of 0.45 M NaOH to 59 mL of the buffer, you need to calculate the final concentration of each specie (the amount of base will increase, while the amount of acid will decrease):

For NaOH:

Ci * Vi = Cf * Vf

0.45M * 0.75mL / 59mL = Cf of NaOH

Final concentration of NaOH in the buffer: 0.00572 M

[HPO₄^-2] = [HPO₄^-2]_initial + 0.00572 M ----> this is the concentration after the addition of NaOH

[H₂PO₄^- ] = [H₂PO₄^- ]_initial - 0.00572 M

If we assume that "0.25 M phosphate buffer" refers to the total amount of phospate present in the buffer:

[HPO₄^-2]_initial + [H₂PO₄^- ]_initial = 0.25M

[HPO₄^-2]_initial / [H₂PO₄^- ]_initial = 2.45

So:  [HPO₄^-2]_initial = 0.25M - [H₂PO₄^- ]_initial

And:   0.25M - [H₂PO₄^- ]_initial  / [H₂PO₄^- ]_initial = 2.45

[H₂PO₄^- ]_initial = 0.0724M and [HPO₄^-2]_initial = 0.178 M

After the addittion of NaOH:

[H₂PO₄^- ] = 0.0724M - 0.00572 M =  0.0667M and [HPO₄^-2] = 0.178 M + 0.00572 M = 0.184 M

And finally:

pH = 7.21 + Log (0.184 M  / 0.0667M  ) = 7.64 the pH increases just a little bit (remember that this is a buffer solution)