Question

Phosphate buffered saline (PBS) is a buffer solution commonly used in biological research. The buffer helps to maintain a constant pH. The osmolality and ion concentrations of the solution usually match those of the human body.

A) You need to prepare a stock solution at pH 7.00 with KH2PO4 and Na2HPO4 (pKa =7.21). What would be the respective concentration of these substances if you wish to obtain the final phosphate concentration [HPO4 −2 ] + [H2PO4 − ] = 0.3 M?

B) What will be the pH of the solution after adding 0.18 L of 0.2 M NaOH?

C) What is the ratio of HPO4 −2 of H2PO4 – is part b, and what is the significance?

D) Tris- HCl buffer is often used as a buffer for biochemical reactions in pH range from 7.1 to 9.1. It is made by dissolving in water Tris (hydroxyl methyl) amino methane, a weak base, and then adding HCl.

The conjugate acid-base pair for Tris is (HOCH2)3 C-NH3 + (here TRISH+ ) and (HOCH2)3 C-NH2 (here TRIS base): its equilibrium is

(HOCH2)3 C-NH3 + ↔ (HOCH2)3 C-NH2 + H+ pKa= 8.08 (at 25 o C)

Assume you have to prepare 200 mL of the buffer. To do so you take 100 mL of 0.1 M Tris base, add HCl and then adjust the volume with pure water.

How many mL of 0.1 M HCl you have to add to it in order to have the final pH of pH = 8.0?

Answer 1

Solution:

A - Given pKa = 7.21
we need to make a stock solution of KH₂PO₄ and Na₂HPO₄ with pH = 7.

As per Handerson Hasselbach equation,
pH = pKa + log ([A-]/[HA])
Here [A-] = [HPO4^-2] and [HA] = [H2PO4^-]
Therefore
7 = 7.1 + log ([A-]/[HA]) => log([A-]/[HA]) = -0.1
[A-]/[HA]= 0.794 => [A-]= 0.794[HA]

Also
[A-]+ [HA]= 0.3M (Given)
0.794[HA] + [HA]= 0.3

[HA] = 0.3/1.794 = 0.167 M
[A-] = 0.794 x 0.167 = 0.1325 M

Therefore, to obtain the final stock solution at pH 7.00 with KH2PO4 and Na2HPO4 , final concentration are [HPO4 −2 ] = 0.132 M and [H2PO4 − ] = 0.167 M.