) Determine the pH of the phosphate buffer containing 100 ml 0,1M KH2PO4 and 100 ml 0,3M Na2HPO4. K (H2PO4-)=1,67*10-7

Answer 1

In this buffer, the weak acid is H₂PO₄^- and the conjugate base is HPO₄^2-

Using the Henderson Hasselbalch equation, we have...

pH = pKa + log [HPO₄^2-] / [H₂PO₄^-]

pKa = -log Ka = -log 1.67x10-7 = 6.78

Since 100 ml of each is mixed with 100 ml of other, the final [ ] of each will be 1/2 the original concentration.

Final [HPO₄^2-] = 0.15 M

Final [H₂PO₄^-] = 0.05 M

pH = 6.78 + log (0.15/0.05) = 6.78 + log 3 = 6.78 + 0.48

pH = 7.26