Download first midterm examination

NAME:__SOLUTION KEY___ Chem 51, fall, 2003, first midterm exam
A periodic table is included as an insert. You may use the back for scratch work but
enter ALL work to be graded in the space provided with each question. Show your work
in all questions involving computation in order to receive credit. All reactions are
conducted in water at 25.00C. Useful data: Student's t at the 95% confidence level for
1, 2, 3, 4, 5 degrees of freedom: 12.707, 4.303, 3.182, 2.776, 2.571.
Ka(acetic acid) = 1.85 x 10-5, Kw = 1.00 x 10-14.
1) (10 points) An analytical laboratory is given the task of determining the concentration
of trichloroethylene (TCE) in a city's water supply. Five samples are taken and the
follow results in parts per million (ppm) were obtained using a very precise gas
chromatograph: 75.07, 74.29, 75.37, 74.52, and 74.77. The mean value and standard
deviation for the data set are 74.804 ppm and 0.429 ppm, respectively.
Provide for a report to the city council, the best value of the concentration and a measure
of its uncertainty at the 95% confidence level to the correct number of significant digits.
The best estimate of the TCE concentration is the mean value. A scatter
plot of the data reveals no obvious outliers so we shall use all of the N = 5
data points. We require the standard deviation of the mean given by s/N0.5
= (0.429 ppm)/50.5 = 0.192 ppm. This statistic shows that the least
significant digit lies in the tenth's place so the TCE concentration should
be reported at 74.8 ppm. We report the uncertainty at the 95% confidence
level. The confidence interval is obtained by multiplying the standard
deviation of the mean by the Student's t. value. With one parameter
extracted from the data (the mean), we have f = N - 1 = 4 and hence t =
2.776. The CI of the mean is (2.776)(0.192 ppm) = 0.53 ppm. Note that
the uncertainty is given to two SD's.
2) (30 points) Write a balanced, net-ionic equation for the reactions that occur when
solutions or slurries of the follow sets of compounds are mixed:
a) iron(II) phosphate and hydrobromic acid
Fe3(PO4)2(s) + 4 H+(aq) = 3 Fe+2(aq) + 2 H2PO4-(aq)
Note. HBr is a strong acid. Phosphoric acid is moderately strong although full
credit was given is it was indicated as undissociated.
b) copper metal and hot, concentrated nitric acid
Cu(s) + 4 H+(aq) + 2 NO3-(aq) = Cu+2(aq) + 2 NO2(g) + 2 H2O(l)
Nitrate ion in the presence of hydronium ion is a strong oxidizing agent at
elevated temperatures. Full credit was given if NO was given as the product.
c) chromium(III) nitrate and potassium cyanide
Cr+3(aq) + 6 CN-(aq) = Cr(CN)6-3(aq)
If only 3 moles of cyanide per mole of chromium were provided, a likely reaction
would be the precipitation of chromium(III) cyanide.
3) (50 points) The weak acid Hexobarbital [C12H16N2O3] is a sedative but it is usually
administered as Privenal [C12H15N2NaO3, MW 258.25 g/mol], the sodium salt of its
conjugate base. The Merck Index reports that the pH of a 10% w/v (0.100 g/ml) solution
of Privenal is 11.5.
a) Hexobarbital has a low solubility in water; Privenal is highly soluble. Explain.
Privenal is an ionic compound and the significant solvation of the anion
and cation greatly contributes to its solutbility in water. A molecule of
Hexobarbital, in contrast, has zero net charge and is relatively non-polar.
Hence, the solute-solvent interactions needed for solubility are weak.
Some students correctly noted that dissociation will not help things as the
substance is a weak acid.
b) Calculate the pKb of the conjugate base Privenal.
This is a weak base problem!
We require first the molar concentration of the base, [B-]0
[B] = (0.1 g/ml)(1000 ml/liter)/(258.25 g/mole) = 0.387 M.
We next require [OH-]
pOH = 14 - pH = 14 - 11.5 = 2.5 so [OH-] = 10-2.5 = 0.0032 M.
For a base B-, the relevant reaction is B- + H20 = BH + OH- so
Kb = [OH-]{BH]/[B-].
This can be simplfied in two ways:
a) Since the sole signficant source of hydroxide is the base, [B-] = [OH-].
b) Since the base is weak, we can replace [B-] with [B-]0 and avoid a
messy quadratic.
Hence Kb = [OH-]2/[B-]0 = (0.0032 M)2/(0.387 M) = 2.6 x 10-5 and pKb =
-log10(Kb) = 4.58.
c) Calculate the pKa of weak acid Hexobarbital.
pKa = pKw - pKa = 14 - 4.58 = 9.42
4) (10 points) Sulfur dioxide is released into the atmosphere when sulfur containing coal
is burned. Explain why this release is a source of acid rain.
Sulfur dioxide is the oxide of a non-metal and therefore will be expected
to react with the water in the atmosphere to yield an acid, the weak acid
H2SO3 in this case. There is another possible fate. SO2 can be slowly
oxidized to form SO3 which upon hydrolysis yields the very strong acid
H2SO4.
5) (20 points) Consider acid rain falling in an area with lakes. Most of the acid rain falls
on the ground and percolates through the soil and porous rock before reaching the lake.
Trout are very sensitive to low pH. Compare the environmental consequences of the acid
rain in two locations. In the first, the predominant rock is limestone (calcium carbonate);
in the other, it is sandstone (fine grains of highly insoluble silicon dioxide).
The acid rain will react with the carbonate anion in the limestone and
liberate gaseous carbon dioxide. The pH of pristine lake water is close to
7 so precipitation of calcium hydroxide requiring a very high hydroxide
concentration will not occur. The calcium enters solution as the Ca+2
cation. As a result, the limestone deposit serves as an enormous acid
buffer and the pH of the lake is not disturbed by the rain.
In contrast, the highly insoluble quartz grains in the sandstone are virtually
inert and no significant reaction occurs with the rain water. As a result the
hydronium ions in the rain inevitably lower the pH of the lake. The lake
eventually will not support trout.
6) (30 points) Calculate the pH of a 250 ml of a solution that is 0.50 M acetic acid and
0.20 M sodium acetate. Recalculate the pH when 0.050 moles of sodium hydroxide is
added to the solution.
This is a buffer problem.
a) Ka = [H+][A-]/[HA]  [H+][A-]0/[HA]0 so [H+]  Ka([HA]0/[A-]0) =
(1.85 x 10-5)(0.50 M/0.20 M) = 4.63 x 10-5, a yielding a pH of 4.33.
b) At first, we have (0.50 M)(0.250 liter) = 0.125 moles of acetic acid and
(0.20 M)(0.250 liter) = 0.050 moles of acetate.
If we add 0.050 moles of hydroxide to the solution (a number less than
0.125 moles), the amount of acetic acid is diminished on a mole-per-mole
base to 0.125 - 0.050 = 0.075 mole. The amount of acetate is at the same
time increase to 0.050 + 0.050 = 0.100 mole. We can convert these
numbers to concentrations but there is no need for this additional step as
the ratio of acetic acid species occurs in the equation and a ratio of
concentrations is the same as the ratio of moles.
[H+] = (1.85 x 10-5)(0.075/0.100) = 1.39 x 10-5 and a pH of 4.86.
ex51_1_03.doc, WES, 25 September 2003