Question

What volume (to the nearest 0.1 mL) of 4.00-M HCl must be added to 0.400 L of 0.150-M K₂HPO₄ to prepare a pH = 6.50 buffer?

mL

Answer 1

K2HPO4 = HPO4-2 present

it is amphoteric, so we must choose either it is a bse/acid

pKa2 = 7.21 is the nearest ionizaiton constant, it acts as a base

so

pH = pKA2 + log(HPO4-2 / H2PO4-) is true

initial HPO4-2 = MV = 0.15*0.4 = 0.06

after adding H+

HPO4-2 = 0.06 - x

H2PO4- = x

6.50 = 7.21 + log( ( 0.06-x) / (x))

10^(6.50 - 7.21) =  (0.06-x) / (x)

0.19498x = 0.06 - x

1.19498x = 0.06

x = 0.06/1.19498

x = 0.05021

we need to add 0.05021mol

V = mol/M = 0.05021 / 4 = 0.0125525 Liters= 12.55 mL