# Chapter 2 Problem Set ```Chap. 2. Problem 2a.
pH = - log [H+]
pH = - log (1.75 x 10-5)
pH = - (-4.76)
pH = 4.76
Chap. 2. Problem 3a.
pH = - log [H+]
[H+] = 10-pH
[H+] = 10-3.82
[H+] = 1.51 x 10-4 M
Chap. 2. Problem 12.
Part a.
Acetic acid ionizes according to the following equilibrium equation
HAc  H+ + AcThe Henderson-Hasselbalch equation for acetic acid is
pH = pKa + log [Ac-]/[HA]
Substituting pKa = 4.76 and 2/1 for [Ac-]/[HA] gives
pH = 4.76 + log (2/1)
pH = 4.76 + 0.30
pH = 5.06
Other answers: (b) 4.28; (c) 5.46; (d) 4.76; (e) 3.76
Chap. 2. Problem 15.
Ionizable proton
A rearranged form of the HH
equation can be used to
determine the fraction of
aspirin in its absorbable
conjugate acid form (HA) as a
function of pH in the stomach
and intestines. In the stomach,
10(pH - pKa) = [A-]/[HA]
10(1.5 - 3.5) = [A-]/[HA]
10-2 = 0.01 = 1/100 = [A]/[HA]
In the small intestine,
10(6 - 3.5) = [A-]/[HA]
102.5 = 316/1 = [A-]/[HA]
The calculations show that a much greater fraction of aspirin is
in its absorbable conjugate acid form in the stomach compared
to the small intestine. Thus more aspirin is absorbed into the
bloodstream from the stomach than from the small intestine.
Chap. 2. Problem 18a&amp;b.
Part (a). A solution of a weak
acid or base is effective in
buffering pH when the pH is &plusmn;
1 unit from the pKa. In this pH
range, appreciable amounts of
both conjugate base and
conjugate acid species are
present in solution. Therefore,
glycine can be used as an
effective buffer in the pH
range of 8.6-to-10.6.
Part (b). This part of the question can be answered using the
HH equation.
pH = pKa + log [R-NH2]/[R-NH3+]
[R-NH2]/[R-NH3+] = 10(9.0-9.6) = 10-0.6 = 0.25 = 1/4
This indicates that the amino group of glycine is about 1/5
deprotonated and 4/5 protonated at pH 9.0.
Chap. 2. Problem 18c.
Part (c). This part of the problem can be answered by first
calculating the number of moles of R-NH2 in 1 L of the 0.1 M
glycine solution at pH 9.0 and 10.0 using the HH equation. The
difference in the number of moles present is stoichiometrically
equivalent to the number of moles of OH- added in the form of KOH.
At pH = 9.0
10(9-9.6) = [R-NH2]/[R-NH3+] = 0.25 = 1/4 = 0.02 moles/0.08 moles.
At pH = 10.0
10(10-9.6) = [R-NH2]/[R-NH3+] = 2.5 = 2.5/1 = 0.0714 moles/0.0286 moles.
During titration from pH 9.0 to 10.0, the number of moles of KOH
0.0714 moles - 0.02 moles = 0.0514 moles KOH.
Thus, the volume of 5 M KOH added is
(0.0514 moles)/(5 moles/L) = 0.01028 L = 10.3 ml KOH
Chap. 2. Problem 18d.
Part (d). This part of the question also can be answered using the
HH equation.
pH = pKa + log [R-NH2]/[R-NH3+]
pH = pKa + log (0.01/0.99)
pH = pKa + (-2) = pKa - 2
In general, a weak acid or base is almost completely protonated
when the solution pH is 2 units below the pKa. Likewise, it is
almost completely deprotonated when the solution pH is 2 units
above the pKa.
Chap. 2. Problem 22.
The equation for the complete ionization of phosphoric acid is
H3PO4  H+ + H2PO4-  H+ + HPO42-  H+ + PO43This problem is solved using the HH equation. Note that the pKa
closest to the pH of the solution should be selected for
substitution into the HH equation. This is because the
concentrations of H3PO4 and PO43- in solution at this pH are
insignificant compared to the concentrations of H2PO4- and HPO42-.
pH = pKa + log [conjugate base]/[conjugate acid]
pH = pKa + log [HPO42-]/[H2PO4-]
[HPO42-]/[H2PO4-] = 10(7.0-6.86)
[HPO42-]/[H2PO4-] = 100.14 = 1.38 = 1.4
Chap. 2. Problem 24.
pH = pKa + log [A-]/[HA]
pH - pKa = log [A-]/[HA]
10(pH - pKa) = [A-]/[HA]
10(5-6) = 10-1 = 0.1 = 1/10 = [A-]/[HA]
Note that this acid is 1/(10+1) = 1/11 = 9.1% ionized
at this pH.
Chap. 2. Problem 28.
pH = pKa + log [conjugate base]/[conjugate acid]
pH = pKa + log [acetate]/[acetic acid]
pH = 4.76 + log (0.2/0.6)
pH = 4.76 + log (0.33)
pH = 4.76 + (-0.48) = 4.28 = 4.3
Chap. 2. Problem 33.
The overall equation for the
carbon dioxide-bicarbonate
buffer system is
CO2 + H2O  H+ + HCO3The pH of blood plasma is
controlled by this system. In
Part (a), during hypoventilation,
as when one holds one’s breath,
the concentration of CO2 in the
blood increases. This shifts the
equation to the right and results
in an increase in the acidity of
blood (lower pH).
In Part (b), during hyperventilation the concentration of CO2 in the
blood decreases. This shifts the carbon dioxide-bicarbonate
equation to the left, decreasing [H+] and raising the pH of the
blood. In Part (c), raising the pH of the blood due to
hyperventilation increases the ability of the blood to resist a
decrease in pH due to the input of H+ into the blood from lactate.
```