CHEMISTRY 51, FALL, 1999, FIRST MIDTERM
NAME: SOLUTIONS

Show your work to receive credit on any question involving computation or derivation. A copy of the periodic table and a table of indicators is included with the examination. You may use the backs for scratch work but enter ALL work to be graded in the space provided with each question. Unless otherwise indicated, all reactions are conducted in water at 25°C at which Kw = 1.00 x 10-14.

1) (30 points) Aspirin, a widely used analgesic, is converted in the body to salicylic acid [MW, 138.12 g/mol] which is responsible for the drug's pharmacological activity. The Merck Index reports that the solubility of salicylic acid in water is 1.0 g per 480 ml and that the pH of a saturated solution is 2.4.

a) Calculate the molar solubility of salicylic acid.

nHA = (1.0 g)/(138.12 g/mol) = 0.0072 mol of salicylic acid, HA
[HA] = (0.0072 mol)/(0.480 l) = 0.015 M

b) Estimate the Ka of salicylic acid. MAKE THE USUAL SIMPLIFYING APPROXIMATIONS!

Assume that [HA}0 ( [HA] (a weak but not outrageous assumption)
stoichiometry: [H+] = [A-] = 10-2.4 = 0.0040 M

Ka = [H+][A-]/[HA] ( [H+]2/[HA]0 = (0.0040 M)2/(0.015 M) = 1.1 x 10-3

c) The conjugate base of salicylic acid is the salicylate anion. Mixing solutions of sodium salicylate and iron(III) nitrate which are colorless or nearly colorless yields a solution that has a deep burgundy color. Account for this striking color change.

Fe(III) is a transition-metal cation and is capable of forming a complex.
The development of the color indicates that this reaction has in fact
occurred. The salicylate anion is a base and hence a ligand. The
-OH group with the lone pairs of electrons can also form a metal-ligand
bond. The salicylate anion is a bidentate ligand, i.e. it forms 2
metal-ligand bonds.

2) (9 points) Chlorine dioxide, ClO2, is used by the city of Zürich to purify its water. What is the chemical basis for its activity as a bactericidal agent? Discuss its use in terms of the four classes of chemical reactions that we examined in class.

The oxidation number of chlorine in chlorine dioxide, +4, is moderately
high and positive so the compound very likely functions as an oxidizing agent.

3) (54 points) The bicarbonate anion, HCO3-, is a weak monoprotic acid with a Ka of 6.0 x 10-11.

a) A chemist wishes to prepare a series of buffers from samples of NaHCO3 [MW, 84.00 g/mol] and Na2CO3 [MW, 106.00 g/mol]. Calculate the pH of the bicarbonate/carbonate buffer with the maximum buffering capacity.

The maximum buffering capacity is reached when [HA} = [A-] whereby
pH = pKa = -log(6 x 10-11) = 10.22

b) The chemist transfers 0.084 g of NaHCO3 and 0.212 g of Na2CO3 to a 250.00 ml volumetric flask and adds water to the mark. Calculate the pH of the resulting buffer.

nHA = (0.084 g)/(84.0 g/mol) = 0.0010 mol
nA- = (0.212 g)/(106.0 g/mol) = 0.0020 mol

Ka = [H+][A-]/[HA] = [H+](nA-/nHA) so [H+] = Ka(nHA/nA-)

[H+] = (6 x 10-11)(0.0010/0.0020) = 3.0 x 10-11

pH = -log(3.0 x 10-11) = 10.52

c) Would the properties of the buffer be changed if potassium carbonate had been used instead of sodium carbonate and the moles of carbonate transferred to the volumetric flask were kept the same? Explain.

The potassium and sodium cations are group Ia cations and are virtually
inert. They are spectator ions and do not influence the reaction.
The substitution of one for the other will not affect the chemistry.

d) Calculate the Kb of the carbonate anion.

Kb = Kw/Ka = (1.0 x 10-14)/(6.0 x 10-11) = 1.7 x 10-4, a small number.

e) The chemist prepares 50 ml of a 0.050 M solution of NaHCO3 and titrates it with 0.025 M NaOH. Calculate the pH at the equivalence point.

At the equivalence point 100 ml of base are added bringing the total volume of
the solution from 50 ml to 150 ml. A dilution factor of 3 results.
Note that the answer to the question is independent of the amount of acid titrated.
The bicarbonate anion is converted to the carbonate anion at the e.p.
This is a weak base problem and due to the dilution effect
[A-] = [HA]0/3 = (0.050 M)/3 = 0.017 M

Kb = [OH-][HA}/[A-} ( [OH-]2/[A-] so [OH-] = [(1.7 x 10-4)(0.017)]0.5 = 0.0017 M
pOH = 2.78 so pH = 14 - pH = 14.00 - 2.78 = 11.22

f) Select the optimum indicator for the titration from the list of indicators provided with the examination.

Find an indicator whose pKa matches the pH at the end point.
Acceptable choices are alizarin and alizarin yellow R.

4) (30 points) A student calibrates a 25 ml transfer pipet and reports the following results: number of determinations, 5; average value, 25.25 ml; standard deviation, 0.025 ml.

a) Determine the uncertainty in the volume of a single use of the pipet at the 95% confidence level. The Student's t value is 12.706, 4.303, 3.182, 2.776, 2.571, 2.447 for 1, 2, 3, 4, 5, and 6 degrees of freedom.

A single use is specified so that we require the 95% C.I. of a single
measurement. CI = ts. We have N = 5 data points and one parameter, the
mean value, and hence 5-1 = 4 degrees of freedom. t = 2.776
and CI = (2.776)(0.025 ml) = 0.069 ml.

b) The student has to use the volume of the solution transferred in later calculations. Which value should be used, one of the 5 measured values or the average? Explain.

Use the best single estimate of the actual value, the average.
Note that averaging reduces random error but does not eliminate systematic error.

c) Do the student's data indicate the presence of systematic error in the manufacture of the pipet that was calibrated? Explain.

The deviation from the nominal value of 25.00 ml, 25.25 - 25.00 = 0.25 ml,
is much larger than the measure of random error, 0.069 ml. Systematic
error, i.e. a flaw, is indicated.

5) (27 points) Write balanced, net-ionic equations for the reactions that occur when aqueous solutions or slurries of the following pairs of reagents are mixed:

a) magnesium metal + acetic acid
Mg is a strong reducing agent.

Mg(s) + 2 CH3COOH(aq) = Mg+2(aq) + H2(g) + 2 CH3COO-(aq)

b) siderite [iron(III) carbonate] + nitric acid

Fe2(CO3)2(s) + 6 H+(aq) = 3 CO2(g) + 2 Fe+3(aq) + 3 H2O(l)

c) manganese(II) hydroxide + potassium cyanide

Mn(OH)2(s) + 6 CN-(aq) = Mn(CN)6-4(aq) + 2 OH-(aq)