Question

Buffer preparation is a necessary part of working with any living cell culture. Buffers are most commonly prepared by starting with a weak acid, HA, or weak base, A-, and adding enough strong acid, H+, or strong base, OH-, to find the correct ratio of [HA]/[A-]. A pH = 7.3 buffer is needed in the lab. This buffer is made by first dissolving 52.25 g K2HPO4 in 400 mL of water. What is the pH of this salt solution? This solution of course will be too basic becuase we only have the base of the buffer present. What must the value of the ratio of [H2PO4-]/[HPO42-] be to create a buffer solution that is pH = 7.3? How many moles of HCl must be added to the 400 mL solution of HPO42-(aq) to prepare the pH = 7.3 buffer solution? How many mL of 1 M HCl must be added to the solution to create the pH = 7.3 buffer solution?

Answer 1

a)

moles of K2HPO4- = 52.25 / 174.17 = 0.3

concentration of HPO42- = 0.3 x 400 / 1000 = 0.12 M

HPO42- + H2O   ----------------> H2PO4- + OH-

0.12                                                   0             0

0.12 - x                                               x             x

Kb2 = x^2 / 0.12 - x

1 x 10^-14 / 6.2×10^–8 = x^2 / 0.12 - x

x = 1.39 x 10^-4

[OH-] = 1.39 x 10^-4 M

pOH = -log [OH-] = -log(1.39 x 10^-4)

         = 3.86

pH = 10.14

pH = pKa2 + log [salt / acid]

7.3 = 7.21 + log [HPO42-]/[H2PO4-]

[HPO42-]/[H2PO4-] = 1.23

[H2PO4-]/[HPO42-] = 0.813