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

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

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

h let me think

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.