2. Amino acids and proteins can act as buffers in the body since they contain fu
ID: 1031438 • Letter: 2
Question
2. Amino acids and proteins can act as buffers in the body since they contain functional groups that may act as weak acids or bases. The side chain of the lysine amino acid has a protonated nitrogen with a Ka of 2.95 x 101. a) If normal blood has a [H301 of 3.98 x 108 M, determine the ratio of neutral (Lys) and protonated ysine (LysH) in blood proteins. b) Under normal blood conditions, would lysine be more likely to act as an acid or base? Explain. c) What is the pH of a 0.90 L buffer solution containing 0.25 mol of NH3 and 0.33 mol of NH CI? If you add 0.050 mol of HCl to adjust the pH, what should the new pH be? (Kb NH3 1.8 x 10-5)Explanation / Answer
2a) Consider the ionization of protonated lysine as below.
LysH+ (aq) + H2O (l) ---------> H3O+ (aq) + Lys (aq); Ka = 2.95*10-11
Given Ka, we have pKa = -log Ka = -log (2.95*10-11) = 10.53.
Again, [H3O+] = 3.98*10-8 M; therefore, pH = -log [H3O+] = -log (3.98*10-8) = 7.40.
Use the Henderson-Hasslebach equation to determine the ratio of the concentrations of neutral Lys and protonated LysH+.
pH = pKa + log [Lys]/[LysH+]
=====> 7.40 = 10.53 + log [Lys]/[LysH+]
=====> -3.13 = log [Lys]/[LysH+]
=====> [Lys]/[LysH+] = antilog (-3.13) = 7.413*10-4 7.41*10-4
The ratio of neutral Lys and protonated LysH+ is 7.41*10-4:1 = (7.41*10-4)/(7.41*10-4):1/(7.41*10-4) = 1:1349.5 1:1350 (ans).
b) Under normal conditions, the pH of blood is about 7.0. The pKa of LysH+ is 10.53. At a pH lower than the pKa of the protonated LysH+, the protonated form of the lysine is the major species while when the pH > pKa, the neutral form is the major species. The protonated LysH+ is an acid and hence, under normal conditions, lysine acts as an acid.
c) The base ionization equation is
NH3 (aq) + H2O (l) --------> NH4+ (aq) + OH- (aq)
Since OH- is a base, we work with pKb. We are given Kb = 1.8*10-5; therefore, pKb = -log Kb = -log (1.8*10-5) = 4.74.
The volume of the solution is constant at 0.90 L; hence, we can express [NH4+]/[NH3] in terms of the number of moles of the components. Therefore, we can write the pOH as
pOH = pKb + log [NH4+]/[NH3]
=====> pOH = 4.74 + log (0.33/0.90 mol/L)/(0.25/0.90 mol/L)
=====> pOH = 4.74 + log (0.33/0.25) = 4.74 + log (1.32) = 4.74 + 0.12 = 4.86.
We know that pH + pOH = 14; therefore, pH = 14 – pOH = 14.00 – 4.86 = 9.14 (ans).
We add 0.050 mole HCl to adjust the buffer. HCl reacts with NH3 to form NH4Cl as
NH3 (aq) + HCl (aq) --------> NH4Cl (aq)
Mole(s) of NH3 neutralized = moles of HCl added = moles of NH4Cl formed = 0.050 mole.
Moles of NH3 retained at equilibrium = (0.25 – 0.050) mole = 0.20 mole.
Moles of NH4+ at equilibrium = (0.33 + 0.050) mole = 0.38 mole.
The volume of the solution is constant at 0.90 L; hence, we can express [NH4+]/[NH3] in terms of the number of moles of the components. Therefore, we can write the pOH as
pOH = pKb + log [NH4+]/[NH3]
=====> pOH = 4.74 + log (0.38/0.90 mol/L)/(0.20/0.90 mol/L)
=====> pOH = 4.74 + log (0.38/0.20) = 4.74 + log (1.90) = 4.74 + 0.279 = 5.019 5.02.
We know that pH + pOH = 14; therefore, pH = 14 – pOH = 14.00 – 5.02 = 8.98 (ans).
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