Blood is a buffered solution with pH values ranging from 7.35-7.45. The importan
ID: 949603 • Letter: B
Question
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. What are the three additional species in solution that can be derived from phosphoric acid using Bronsted-Lowry acid-base concepts? (I will refer to them as Species #1, #2, and #3 in the next parts.) 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.) For a pH value of 7.40, calculate the following ratios (you may use relevant K_a values that are tabulated for 25 degree C in Appendix D of your textbook): Which forms of phosphoric acid are the most abundant and the least abundant in water at a pH of 7.40? 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 WCI(aq) stock solution, (6) a 3.00 M NaOH(ag) stock solution, and (7) deionized water. To make an optimum buffer, it is important that the ipK_a of the acid part is close to the target pH of the buffer. Which phosphate species has a pK_a value closest to pH = 7.35? 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? 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?Explanation / Answer
a. H2PO4- , HPO42- , PO43-.
b. 1. H3PO4(s) + H2O(l) <---> H3O+(aq) + H2PO4(aq) ; Ka1= 7.5×103 ; pKa = 2.12
2. H2PO4(aq) + H2O(l) <---> H3O+(aq) + HPO42(aq) ; Ka2= 6.3 ×108 ; pKa = 7.2
3. HPO42(aq) + H2O(l) <---> H3O+(aq) + PO43(aq) ; Ka3= 4.2×1013 ; pKa = 12.4
c. Henderson equation, pH = pKa + log[salt]/[acid]
1. 7.4 = -log(7.5×103) + log [H2PO4]/[ H3PO4]
or, [H2PO4]/[ H3PO4] = 188391
2. 7.4 = -log(6.3 ×108) + log [HPO42]/[ H2PO4]
or, [HPO42]/[ H2PO4] = 1.58
or, [HPO42]/[ H2PO4] * [H2PO4]/[ H3PO4] = 188391 * 1.58
or, [HPO42]/[ H3PO4] = 298126
3. 7.4 = -log(4.2×1013) + log [PO43]/[ HPO42]
or, [PO43]/[ HPO42] = 1.05*10-5
or, [PO43]/[ HPO42] * [HPO42]/[ H3PO4] = 298126 * 1.05*10-5
or, [PO43]/[ H3PO4] = 3.14
d. [HPO42]/[ H3PO4] > [H2PO4]/[ H3PO4] > [PO43]/[ H3PO4] at pH = 7.4
So, HPO42 is the most abundant and PO43 is the least abundant species at pH = 7.4
e . i. H2PO4(aq) + H2O(l) <---> H3O+(aq) + HPO42(aq) ; Ka2= 6.3 ×108 ; pKa = 7.2
Ans: HPO42
ii.
H3O+(aq) + HPO42(aq) <---> H2PO4(aq) + H2O(l)
pH = pKb + log [salt]/[base]
or, 7.35 = (14-7.2) + log [H2PO4]/[ HPO42]
[H2PO4]/[ HPO42] = 3.548 M = 3.548 moles (as solution is 1 L)
Given, [H2PO4] + [ HPO42] = 0.15 = 0.15 moles
solving we get, H2PO4 = 0.117 moles
HPO42 = 0.033 moles
To a total 0.15 moles of [ HPO42] we have to add x mL 3M HCl so that the final solution is 1 L and pH = 7.35
As H2PO4 = 0.117 moles ,
so HCl added = 0.117 moles = 0.117*1000/3 or 39 mL 3 M HCl.
Initial HPO42 = 0.15 moles
this 0.15 moles HPO42 is present in (1000-39)or 961 mL HPO42 solution.
Strength of HPO42 solution = 0.15*1000/961 = 0.156 M
961 mL 0.156 M HPO42- solution and 39 mL 3M HCl solution have to be mixed.
prepare 961 mL 0.156 M HPO42- by dissolving (M*961*0156/1000) g of sodium monohydrogen phosphate in 961 mL water; where M = molar mass of sodium monohydrogen phosphate.