Question

How many milliliters of 4.7 M HCl must be added to 3.9 L of 0.11 M K₂HPO₄ to prepare a pH = 7.44 buffer?

Answer 1

K₂HPO₄ MOLES= 3.9 x0.11= 0.429

addition HCl to K₂HPO₄ results in a buffer solution when HCl is added in lesser amount than K₂HPO₄

K₂HPO₄ + HCl ----------------------> KH2PO4 + KCl

0.429          a                                     0            0

0.429-a        0                                   a              a

thus buffer solution contain 'a' moles of KH2PO4 and (0.429-a) moles K₂HPO₄

pKa = -log Ka = -log(6.2 x10^-8) = 7.20

For acidic buffer

Henderson-Hasselbalch equation

pH = pKa + log[salt/acid]

7.44 = 7.20 +log[salt/acid]

0.24 = log[(0.429-a)/a]

[(0.429-a)/a] = 10^0.24= 1.737

0.429 -a = 1.737a

0.429= 2.737a

a= 0.157 mole

M x volume = 0.157

4.7 x V= 0.157

V= 0.0334 L

Volume = 33.4ml of HCl must be added