Question

Consider the following four titrations (a-d):

a. 120 mL of 0.2 M C2H5NH2 (Kb = 5.6×10-4) by 0.2 M HNO3
b. 120 mL of 0.2 M KOH by 0.2 M HNO3
c. 120 mL of 0.2 M HC3H5O2 (Ka = 1.3×10-3) by 0.2 M KOH
d. 120 mL of 0.2 M HNO2 (Ka = 4.0×10-4) by 0.2 M KOH

Part 1
Rank the four titrations in order of increasing pH at the halfway point to equivalence (lowest to highest pH). Use the letters a, b, c, and d to represent the various titrations.



Part 2
Rank the four titrations in order of increasing pH at the equivalence point. Use the letters a, b, c, and d to represent the various titrations.



Part 3
Which titration requires the largest volume of titrant (HNO3 or KOH) to reach the equivalence point? (Enter "none" if no titration requires more than the others.)

Answer 1

Part 1:
The halfway point to equivalence corresponds to the point where half of the acid or base has reacted with the titrant. At this point, the moles of acid and base are equal and the solution is a buffer solution. The pH of a buffer solution depends on the pKa of the acid and the ratio of the concentrations of the acid and its conjugate base (or the base and its conjugate acid). Using the Henderson-Hasselbalch equation, we can estimate the pH at the halfway point for each titration:

a. C2H5NH2 + HNO3 → C2H5NH3+NO3-
pKa = 10.74
At the halfway point, [C2H5NH2] = [C2H5NH3+], so
pH = pKa + log([C2H5NH3+]/[C2H5NH2]) = 10.74 + log(1) = 10.74
b. KOH + HNO3 → KNO3 + H2O
At the halfway point, [HNO3] = [KNO3], so the solution is neutral and pH = 7.
c. HC3H5O2 + KOH → KC3H5O2 + H2O
pKa = 4.87
At the halfway point, [HC3H5O2] = [C3H5O2-], so
pH = pKa + log([C3H5O2-]/[HC3H5O2]) = 4.87 + log(1) = 4.87
d. HNO2 + KOH → KNO2 + H2O
pKa = 3.30
At the halfway point, [HNO2] = [NO2-], so
pH = pKa + log([NO2-]/[HNO2]) = 3.30 + log(1) = 3.30

Ranking from lowest to highest pH at halfway point: b, a, c, d (Note that titration b is a strong acid-strong base titration, which does not form a buffer and has a neutral pH at the halfway point.)

Part 2:
At the equivalence point, all of the acid or base has reacted with the titrant and the solution is either neutral or basic, depending on the acid-base strength of the components. For a strong acid-strong base titration, the equivalence point is pH 7. For a weak acid-strong base titration, the equivalence point is greater than pH 7, and for a strong acid-weak base titration, the equivalence point is less than pH 7. Using this information, we can rank the titrations at the equivalence point:

a. pH at equivalence point < 7 (weak base-strong acid titration)
b. pH at equivalence point = 7 (strong acid-strong base titration)
c. pH at equivalence point > 7 (weak acid-strong base titration)
d. pH at equivalence point < 7 (weak acid-strong base titration)

Ranking from lowest to highest pH at equivalence point: a, d, b, c

Part 3:
The volume of titrant required to reach the equivalence point depends on the number of moles of acid or base present in the sample. For a 0.2 M solution of acid or base, the number of moles present in 120 mL is:

moles = concentration x volume = 0.2 x 0.12 = 0.024

Using this information, we can calculate the number of moles