Question

Calculate the pH of a buffered solution containing 0.040M NaH2PO4 and 0.090M Na2HPO4. pKa's of phosphate buffer : pk1 = 2.10, pk2 = 7.20, pK3=12.40 To answer this question we only use the value of pk2 , why is that? Why not pk1 or pk3? pls help

Answer 1

H3PO4   <--------------->   H+ +    H2PO42-     ,   Ka1 = [H2PO42-] / [H3PO4]

H2PO42-   <------------>   H+   +   HPO42-         , Ka2 = [HPO42-] / [H2PO4-]

HPO42-   <--------------> H+    + PO43-           , Ka3 = [HPO42-] / [PO43-]

here we have :

0.040M NaH2PO4 (H2PO4-) and 0.090M Na2HPO4 (HPO42-)

0.040 M H2PO4- , 0.090 M HPO42-

so here we have to use 2nd Ka value of H3PO4.

pH = pKa2 + log [conjugate base / acid]

     = 7.20 + log [0.090 / 0.040]

pH =7.55