Blood is a buffered solution with pH values ranging from 7.35-7.45. The importan
ID: 949346 • 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. ( there will be three species = H2PO4- , HPO42- , PO43-
Ka1 = 7.25 * 10^-3
Ka2 = 6.31 * 10^-8
Ka3 = 3.98 * 10^-13
1) Write the three equilibria that relate phosphoric acid with each of these species listed. (You should write three chemical equilibria, using chemical symbols, that all resemble the following.)
2)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]/[Phosphoric Acid]
[Species #2] / [Phosphoric Acid]
[Species #3]/ [Phosphoric Acid]
3) Which forms of phosphoric acid are the most abundant and the least abundant in water at a pH of 7.40?
For question# 2 I got
a) 182161
b) 288803
c) 2.888
Explanation / Answer
a.
H3PO4 = H2PO4- + H+
H2PO4- = HPO42- + H+
HPO42- = PO43- + H+
b.
Ka1 = 7.25 * 10^-3 pKa1 = 2.14
Ka2 = 6.31 * 10^-8 pKa2 = 7.20
Ka3 = 3.98 * 10^-13 pKa3 = 12.40
From the buffer formula :
log([base]/[acid]) = pH – pKa
log ([H2PO4-]/[ H3PO4]) = pH – pKa1 = 7.40-2.14 = 5.26
log ([HPO42-]/[ H2PO4-]) = pH – pKa2 = 7.40-7.20 = 0.20
log ([PO43-]/[ HPO42-]) = pH – pKa3 = 7.40-12.40 = - 5.00
[H2PO4-]/[ H3PO4] = 105.26 = 182 000 : 1
[HPO42-]/[ H2PO4-] = 100.20 = 1.58 : 1
[PO43-]/[ HPO42-] = 10-5.00 = 1: 100 000
Thus
[HPO42-]/[ H3PO4] = 182 000 x 1.58 = 284 000 : 1
[PO43-] / [ H3PO4] = 182 000 x 1.58 /100 000 = 2.84
Your results are correct but keep only 3 significant figures. (as in Ka)
c. see the previous results:
HPO42- is the most abundant
H3PO4 is the least abundant