Question

what is the pH of a buffer system by dissolving 17.42g of KH2PO4 and 20.41g of K2HPO4 in water to give a volume of 200ml

Answer 1

Here we have a buffer in which acid is H₂PO₄^- and its conjugate base HPO₄^2-

We also know the [KH₂PO₄ ]= [H₂PO₄^-] since salt dissociates completely. Same is true for

K₂HPO₄ .

Lets calculate their concentrations.

[H₂PO₄^-] =[KH₂PO₄ ]= mol / Volume of solution in L

= (mass in g /molar mass of KH₂PO₄ ) / 0.200 L

= ( 17.42 g / 136.0838 g per mol)/0.200 L = 0.64 M

[HPO₄^2-] =[K₂ HPO₄ ]= (20.41 g / 174.1739 g per mol)/0.200 L

=0.586 M

We will need the ka value of acid H₂PO₄^- that is nothing but the ka2 of phosphoric acid H₃PO₄

ka ₂ (phosphoric acid) = 6.2 E-8

calculation of pka

pka = - log ka = - log (6.2E-8) = 7.21

Lets use Henderson hasselbalch equation

pH = pka + log ([conj. Base ] /[weak acid] )

here conj. Base = HPO₄^2- and weak acid is H₂PO₄^-.

Lets plug all the values in above equation and get the pH.

pH = 7.21 + log ( 0.586 /0.64)

pH = 7.17

pH of this buffer would be 7.17