Solution Key, Homework 3
Problem 1: The volume of the HCl solution is calculated from the usual formula (see
HWK1):
V = {FW(HCl) × mol(HCl) × 100}/{(%HCl) × density}
Hence the errors contributing to the standard deviation of V are as follows:
s2rel(V) = s2rel(mol) + s2rel(density) + s2rel(%) = (0.001/0.05)2 + (0.01/1.18)2 + (0.5/37.0)2 =
0.0006544
srel(V) = 0.02558; s(V) = 0.02558 × 4.128 = x = 0.106 mL
Problem 2: At equivalence point: n(NH3) in sample = n(HCl) = c(HCl) × V(HCl) =
0.1036 mol L-1 × 14.22 mL = 1.47319 mmol
mg(NH3) = mmol × FW = 1.47318 × 17.0307 = 25.0895 mg
Now convert from aliquot to sample: mg(NH3) = 25.0895 × g(sample)/g(aliquot) =
(10.231 + 39.466)/4.373 = 285.13 mg.
%(NH3) = 100 × (0.28513/10.231) = 2.787% (report 4 significant figures)
Problem 3: Calculate % NH3 from the five measurements as explained for Problem 2, then
calculate the mean and standard deviation of the %.
Problem 4: From the definition of dissociation fraction for an acid:
αA = [A]/cHA where [A] is the equilibrium concentration of the conjugate
base, and cHA is the total (formal) concentration of the acid, e.g. cHA = [HA] + [A] for a
monoprotic acid.
From equilibrium: αA = Ka/{Ka + [H3O+]} = Ka/{Ka + 10-pH}
Since Kb is given for cocaine free base, you need to calculate KBH = Ka of the conjugate acid
(cocaine hydrochloride). KBH = 10-14/Kb = 3.85 × 10-9.
Hence: αA = 3.85 × 10-9/{3.85 × 10-9 + 10-7.4} = 0.0882 or 8.8%. Note that you do not need to
know the amount of the cocaine used. As long as the pH remains constant, the fraction of
cocaine hydrochloride that dissociates to the free base is constant. The amount of the free base is
of course proportional to the amount of cocaine hydrochloride added to the sample.
Problem 5: Check the textbook for these exercises.