1. For biological and medical applications we often need to consider proton tran
ID: 969630 • Letter: 1
Question
1. For biological and medical applications we often need to consider proton transfer equilibria at body temperature (37° C). The value of Kw for water at body temperature is 2.5 × 10-14. (a) What is the value of [H3O+] and the pH of neutral water at 37° C? (b) What is the molar concentration of OH- ions and the pOH of neutral water at 37° C?
2.Numerous acidic species are found in living systems. Write the proton transfer equilibria for the following biochemically important acids in aqueous solution: (a) H2PO4- (dihydrogenphosphate ion), (b) lactic acid (CH3CHOHCOOH), (c) glutamic acid (HOOCCH2CH2CH(NH2)COOH), (d) glycine (NH2CH2COOH), and (e) oxalic acid (HOOCCOOH).
3. Suppose that something had gone wrong in the Big Bang, and instead of ordinary hydrogen there was an abundance of deuterium in the Universe. There would be many subtle changes in equilibria, particularly the deuteron transfer equilibria of heavy atoms and bases. The Kw for D2O, heavy water, at 25° C is 1.35 × 10-15. (a) Write the chemical equation for the autoprotolysis (more precisely, autodeuterolysis) of D2O. (b) Evaluate pKw for D2O at 25° C. (c) Calculate the molar concentrations of D3O+ and OD- in neutral heavy water at 25° C. (d) Evaluate the pD and pOD of neutral heavy water at 25° C. (e) Formulate the relation between pD, pOD, and pKw (D2O).
Explanation / Answer
1) Kw = [H+][OH-]
2.5 X 10^-14 = [H+][OH-]
for neutral water
[H+] = [OH-] = 1.58 X 10^-7
pH = -log [H+] = 6.8
pOH = 6.8
2) (a) H2PO4- (dihydrogenphosphate ion),
HPO4- + H2O -> H3O+ + HPO4-2
HPO4-2 + H2O --> PO4-3 + H3O+
(b) lactic acid (CH3CHOHCOOH),
CH3CHOHCOOH + H2O --> CH3CHOHCOO- + H3O+
(c) glutamic acid (HOOCCH2CH2CH(NH2)COOH),
HOOCCH2CH2CH(NH2)COOH + H2O --> HOOCCH2CH2CH(NH2)COO- + H3O+
(d) glycine (NH2CH2COOH),
NH2CH2COOH + H2O --> NH2CH2COO- + H3O+
(e) oxalic acid (HOOCCOOH).
HOOCCOOH + H2O --> HOOCCOO- + H3O+
HOOCCOO- + H2O --> -OOCCOO- + H3O+
3)
a) D2O + D2O --> D3O+ + OD-
b) pKw = -logKw = -log 1.35 X 10^-15 = 14.87
c) The molar concentration will be
D3O+ = OD- = (Kw)1/2 = 0.367 X 10^-7
d) pD = -logD+ = 7.43
pOD = -log[OD-] = 7.43
e) pKw = pD + pOD
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