triprotic acid and buffer

Question

phosphoric acid (h₃po₄) is a triprotic acid with pka values of 2.12, 7.21, and 12.32. it is a useful buffer solution for ph values near the pka values. you have a 0.100 m solution of phosphate buffer with a ph of 2.50. how many ml of 0.500 m naoh solution must be added to 1.00 l of this buffer solution to convert it to a buffer at ph 7.70?

Answer 1

This problem is dealing with the first 2 dissociations of phosphoric acid, as follows:

H₃PO₄ ==> H⁺ + H₂PO₄^- pKa1 = 2.12

H₂PO₄^- ==> H⁺ + HPO₄^2- pKa2 = 7.21

We can use the Henderson Hasselbalch equation to find the answers to this question.

HH equation: pH = pKa + log [conj.base]/[acid]

First, we can find the conditions of the original buffer. This will be a combination of H₃PO₄ (weak acid) and H₂PO₄^- (conjugate base).

2.50 = 2.12 + log (H₂PO₄^-/0.1)

[H₂PO₄^-] = 0.240 M

So original buffer contains 0.1 M H₃PO₄ and 0.240 M H₂PO₄^-

Now, we will add NaOH, which will react with the H₃PO₄ to convert it to H₂PO₄ and then additional to convert H₂PO₄ to HPO₄

H₃PO₄ + OH- ==> H₂PO₄

H₂PO₄ = OH- ==> HPO₄^2-

pH = pKa + log [HPO₄^2-]/[H₂PO₄^-]

7.7 = 7.21 + log [HPO₄^2-]/[H₂PO₄-]

[HPO₄^2-]/[H₂PO₄-] = 3.1 meaning we need a ratio of 3.1 HPO₄ : H₂PO₄

To convert 0.1 mol H₃PO₄ to 0.1 mol H₂PO₄ requires 0.2 L 0.5 M NaOH

To convert 0.1 mol H₂PO₄ to HPO₄ so ratio is 3.1 : 1 requires 0.151 L NaOH**

Total volume of 0.5 M NaOH = 0.2 L + 0.151 L = 0.351 L = 351 mls of 0.5 M NaOH

** To calculate this value we solve to obtain a ratio of 3:1 as follows:

Let x = [HPO₄^2-], then x/0.1-x = 3.1 and x = 0.0756

To get 0.0756 moles of HPO₄^2- using 0.5 M NaOH, you have (x L)(0.5 mol/L) = 0.0756 and x = 0.151 L