Assume an activity coefficient of 1 for all substances and n

Assume an activity coefficient of 1 for all substances and no effect of ionic strength. Eliminate terms in quadratic solutions for [H+ ] only if the weak acid is dissociated < 5%. Reported pKa values can vary depending on the conditions under which they were measured; therefore, in solving the following problems use the pKa values given with the problems.

Phosphate is the most important buffer in cells. Consequently, Phosphate Buffered Saline (PBS) is a commonly used buffer in the biochemistry lab. The normal composition of PBS (1X) is:

a. What is the pH of an 11.8 mM solution of H3PO4? (For H3PO4, pKa1 = 2.12, pKa2 = 7.21, pKa3 = 12.32) Use Ka equation. Assume only dissociation of protons with pKa1, since they have the least affinity. Once pH is calculated, it can be compared to pKa2 value to see if dissociation of protons with pKa2 is significant and needs to be considered in the calculation.

b. What is the pH of PBS?

c. Why does PBS contain NaCl and KCl, and why is their total amount about 140 mM?

All steps to the solutions of the following problems must be shown. Round your answers as appropriate, but no more than three significant figures.

Salt Concentration (mmol/L) Concentration (g/L) 137 8.0 NaCl 0.2 KCI 2.7 10 1.44 Na2HPO4 0.24 1.8 KH2PO4

Solution

1. To calculate pH of 11.8 mM H3PO4 we use following steps

Change mM into M = 11.8mM = 11.8 x 10^-3M = 0.000118 M

pKa1 = - logKa1

Ka1 = antilog (-pKa1)= antilog (-2.12) = 0.0075 = 7.75 x 10^-3

If we count only PKa1 then use the formula to get Kb value

Kw = Ka x Kb

Where Kw is ionization coefficient of water and Ka = Ka1 is dissociation coefficient for weak acid and Kb is dissociation coefficient for weak base

Kw = 1 x 10^-14 always

1 x 10^-14 = 7.75 x 10^-3 x Kb

Kb = 1 x 10^-14 / 7.75 x 10^-3 = 0.129 x 10^-11 = 1.29 x 10^-10

For a weak base A^- + H₂O = HA + OH^-

Kb = [HA] [OH^-]/ A^- ; in which a concentration of A- will dissociate into ‘a’ concentration for both HA and OH. But here H3PO4 is a weak acid so 0.000118- a will dissociate

Now Kb = a x a/ (0.000118 –a)

a² = Kb / 0.000118 [neglect a here] = 1.29 x 10^-10 / 11.8 x 10^-3= 1.093 x 10^-7 = 10.93 x 10^-8

a² = 10.93 x 10^-8

a = 10.93 x 10^-8 = 3.30 x 10^-4

The ‘a’ concentration is also for OH^- concentration so

pH = - log [OH^-]

pH = - log[3.30 x 10^-4 ]= -[ log3.30 -4log10] = -[0.519 – 4] = -[-3.481]= 3.481

Find Ka 2 value =

Ka2 = antilog (-pKa2)= antilog (-7.21) = 1.61 x 10 ^-8 which is much lower than Ka1 and hence neglected.

2. To calculate pH of PBS, according the Handerson-Hasselbach equation, we have to have the value of pKas of Phosphoric acid. In addition, we know the pKa2 value which is 7.21. here we take that value.

We have concentrations of weak base Na₂HPO₄ and KH₂PO₄ which is a conjugate acid and

pH= PKA + log [A^-/HA]

= 7.21 + log[10/1.8] = 7.21 + log 10- log1.8 = 7.21+ 0.988 = 8.198