Question

Copy of A buffer consists of 0.33 M Na2HPO4 and 0.28 M Na3PO4. Given that the K values for H3PO4 are, Ka1 = 7.2 x 10-3, Ka2 = 6.3 x 10-8, and Ka3 = 4.2 x 10-13, calculate the pH for this buffer

Answer 1

Use Ka3

henderson-hasselbach equation: pH = pKa + log [base]/[acid]

pKa3 = - log Ka3 = - log 4.2 x 10^-13 = 12.38

therefore: pH = 12.38 + log (0.28/0.33) = 12.30

the pH is 12.30

Step-by-step explanation:

phosphoric acid is a polyprotic acid meaning it donates more than one proton

weak Acid ↔ conjugate Base

H3PO4 ↔ H2PO4^- corresponding to Ka1

H2PO4^- ↔ HPO4^2- corresponding to Ka2

HPO4^2- ↔ PO4^3- corresponding to Ka3

A buffer consist of a weak acid and its conjugate base, the given buffer has the combination of HPO4^2- and PO4^3- thud we used Ka3

knowing that we used henderson-hasselbach equation to get the pH which is 12.30