In: Chemistry
Blood is a buffered solution with pH values ranging from 7.35-7.45. The important components of blood that contribute to its buffering capacity are species derived from carbonic acid and phosphoric acid. This problem emphasizes the phosphoric acid component.
(a) (3 pts) What are the three additional species in solution that can be derived from phosphoric acid using Brønsted-Lowry acid-base concepts? (I will refer to them as Species #1, #2, and #3 in the next parts.)
(b) (3 pts) Write the three equilibria that relate phosphoric acid with each of these species listed in part (a). (You should write three chemical equilibria, using chemical symbols, that all resemble the following.) “Phosphoric Acid” + ___________ “Species #” + ___________ (The items in blanks are needed to properly balance the equations.)
(c) (3 pts) For a pH value of 7.40, calculate the following ratios (you may use relevant Ka values that are tabulated for 25C in Appendix D of your textbook): [Species #1] [Species #2] [Species #3] ; ; [Phosphoric Acid] [Phosphoric Acid] [Phosphoric Acid]
(d) (2 pts) Which forms of phosphoric acid are the most abundant and the least abundant in water at a pH of 7.40?
(e) (10 pts) You work in a medical research lab with a biochemist who asks you to make a 1.00 L phosphate buffer stock solution with pH = 7.35. The total phosphorus concentration is to be 0.150 M. In the storeroom, there are the following reagents: (1) a 3.00 M phosphoric acid stock solution, (2) solid sodium dihydrogen phosphate, (3) solid sodium monohydrogen phosphate, (4) solid, sodium phosphate, (5) a 3.00 M HCl(aq) stock solution, (6) a 3.00 M NaOH(aq) stock solution, and (7) deionized water.
(i) (2 pts) To make an optimum buffer, it is important that the pKa of the acid part is close to the target pH of the buffer. Which phosphate species has a pKa value closest to pH = 7.35?
(ii) (4 pts) Choose this phosphate species (in i) as the only phosphorus component. What amounts of this substance and any other reagents (grams of solid(s); mL of any solution(s) and water) would you need to make the buffer with pH = 7.35?
(iii)(4 pts) Choose this phosphate species (in i) and either its conjugate acid or base as the only two phosphorus components. What amounts of this substance, its conjugate acid or base, and water (grams of solid(s); mL of any solution(s) and water) would you need to make the buffer with pH = 7.35?
You have multiple questions in a same post, I will answer some of them, and give you tips for answering the rest of the question.
First, the H3PO4 is a triprotic acid, and these 3 protons can be dissociated in water to give three differents species which are:
H2PO4-(1), HPO42-(2), PO43-(3)
For part b) the equilibrium reactions (along with the Ka values)
are:
H3PO4 + H2O -------->
H2PO4- +
H3O+ Ka1 = 7.1x10-3
---> pKa1 = 2.15
H2PO4- + H2O
--------> HPO42- +
H3O+ Ka2 = 6.2x10-8
---> pKa2 = 7.21
HPO42- + H2O -------->
PO43- +
H3O+ Ka3 = 4.6x10-13
--> pKa3 = 12.34
For part c) I don't know very well this part of the ratio you are mentioning because it does not see well, what are the ratios, however, let me tell you how you'll do it. To calculate the ratio for any concentration, you need to use the HH equation which is:
pH = pKa + log([A-] / [HA] The ratio is
[A-]/[HA] so, you have to solve for that in the equation
which is:
[A-]/[HA] = 10pH-pKa
So, let's do a calculation example for the first ratio between phosphoric acid and species 1:
[H2PO4-]/[H3PO4]
= 107.40-2.15
[H2PO4-]/[H3PO4]
= 177,827.94
Now, with this calculation example, do the other ratios.
For part d) at that pH the most abundant form will be the one with the pKa closer to the value of pH desired. In this case, at pH = 7.40, the species with a pKa closer to that value is the second pKa of 7.21, that belongs to the HPO42-. The least abundant will be the one with the farthest value of that pH. And in this case is the pKa1 belonging to H3PO4.
Now for part e) post that in another question thread. However for that pH desired, choose the buffer that has the relation between HPO4 and H2PO4 (pKa = 7.21).
Hope this helps