Question

How would you prepare 1 L of a 0.100 M phosphate buffer at pH 7.02 using crystalline Na₂HPO₄ and a solution of 1.0 M HCl?

Answer 1

You will need the pKa2

pKa2 = 7.21

then

M = 0.1

pH = pKa + log(A-/HA)

7.02 = 7.21 + log(A-/HA)

solve for ratio

ratio= 10^(7.02-7.21) = 0.64565

conjugate / acid = 0.64565

if the reaction is

H2PO4- <--> HPO4-2 + H+

then

acid = H2PO4-

base = HPO4-2

0.64565(H2PO4-) = (HPO4-2)

Assume then

H2PO4 = M + 1*V =

HPO4 = M - 1*V

these are concentration so

0.64565(M + V) = (M - V)

M= 0.1 and

V

change to mL

0.64565(0.1 + V/1000) = (0.1 - V/1000)

0.64565 + 0.64565/1000 * V = 0.1 - V/1000

V(0.64565/1000+1/1000) = 0.1 -0.64565

V =( 0.1 -0.64565 )/((0.64565/1000+1/1000) = 331.571111 ml

You must add V = 0.3 L of acid to achieve that pH