Phosphoric acid H3PO4 has three dissociable protons (it is triprotic) thus three pKa values; 2.12, 7.21, and 12.32). What volume of 200 mM phosphoric acid and w

Question

Phosphoric acid H3PO4 has three dissociable protons (it is triprotic) thus three pKa
values; 2.12, 7.21, and 12.32). What volume of 200 mM phosphoric acid and what
volume of 500 mM NaOH are required to make 100 mL of 20 mM phosphate buffer
pH 12?

Answer 1

A phosphate buffer at pH 12 would not be very useful in practice, but as this is probably just an exercise, here is an abbreviated explanation. Hopefully, you can complete it on your own.

 

H₃PO₄ ==> H⁺ + H₂PO₄^-   pKa = 2.12

H₂PO₄^- ==> H⁺ + HPO₄^2-  pKa = 7.21
HPO₄^2- ==> H⁺ + PO₄^3- pKa = 12.32

 

For a buffer at pH 12 (if you actually wanted to do that), we would want to use a solution of HPO₄^2- and PO₄^3- since the pKa of 12.32 is close to the desired pH. That is, the weak acid would be Na₂HPO4 and the conjugate base would be Na₃PO₄

Using the Henderson Hasselbalch equation, we can find the ratio of these two species:

pH = pKa + log [conj.base]/[weak acid]

12 = 12.32 + log [Na₃PO4]/[Na₂HPO4]

log [Na₃PO_4]/[Na₂HPO₄] = -0.32

[Na₃PO₄]/[Na₂HPO₄] = 0.479

 

Recalling that the weak acid is Na2HPO4 and the conjugate base is Na3PO4, proceed as follows and finish the calculations:

Calc. decimal fraction of each:

conj. base = 0.479/0.1479 = 0.3239 

acid = 0.100/0.1479 = 0.6761

 

Final molarity of each:

conj. base = 0.02 M x 0.3239 = 0.006478 M

acid = 0.02 M x 0.6761 = 0.01352 M

 

Find moles of each:

conj. base = 0.006478 mol/L x 0.1 L = 6.478x10^-4 moles

acid = 001352 mol/L x 0.1 L = 1.352x10^-3 moles

 

From H₃PO₄ + 3NaOH ==> Na₃PO₄ + 3H₂O, hopefully you can finish the calculations of mls of NaOH and mls of H₃PO₄ to get the desired moles of conjugate base and acid.