Calculate the pH of a solution of 15 mL K2HPO4 at 01 molarit

Calculate the pH of a solution of 15 mL K2HPO4 at 0.1 molarity, 5 mL KH2PO4 at 0.1 molarity and 30 mL water.

As it is is buffer soln

so pH= pKa+log[K2HPO4]/[KH2PO4]

= 7.16+log[1.5]/0.5

= 7.637

As we knw

Ka for Kh2po4 is 6.2 *10^-8

and Pka= -logka

[k2hpo4] = 15*0.1 = 1.5 mM

and [Kh2po4] = 0.1*5 = 0.5