Download for the exam on 14 feb

Suggested Problems
For the week ending M 11 Feb 08
16.19 Consider the titration of 40.0 mL of a 0.0250 M solution of the protonated form of
the amino acid valine (H2A+) with 0.100 M NaOH. Calculate the pH after addition of
the following volumes of 0.100 M NaOH:
a. 10.0 mL
b. 15.0 mL
c. 20.0 mL
16.92 If a saturated solution prepared by dissolving Ag2CO3 in water has [Ag+] = 2.56 *
10-4 M, what is the value of Ksp for Ag2CO3?
16.96 Use Le Chatelier’s Principle to explain the following changes in the solubility of
Ag2CO3 in water:
a. Decrease on addition of AgNO3
b. Increase on addition of HNO3
c. Decrease on addition of Na2CO3
d. Increase on addition of NH3
16.104 Write a balanced net ionic equation for each of the following dissolution
reactions, and use the appropriate Ksp and Kf values in Appendix C to calculate the
equilibrium values for each:
a. AgI in aqueous NaCN to form Ag(CN)216.108 Will a precipitate of BaSO4 form when 100 mL of 4.0 * 10-3 M BaCl2 and 300
mL of 6.0 * 10-4 M Na2SO4 are mixed? Explain.
16.109 Will a precipitate of PbCl2 form on mixing equal volumes of 0.010 M Pb(NO3)2
and 0.010 M HCl? Explain. What minimum Cl- concentration is required to begin
precipitation of PbCl2 from 5.0 * 10-3 M Pb(NO3)2?
16.112 The pH of a sample of hard water having [Mg2+] = 2.5 * 10-4 M is adjusted to pH
10.80. Will Mg(OH)2 precipitate?
16.131 What is the molar solubility of Mg(OH)2 in a buffer solution that has a pH of
9.00?
17.8 The following data apply to the vaporization of mercury: Hvap = 58.5 kJ/mol;
Svap = 92.9 J/(K*mol)
a. Does mercury boil at 325oC and 1 atm pressure?
b. What is the normal boiling point of mercury?
17.38 Predict the sign of the entropy change in the system for each of the following
processes:
a. A solid sublimes
b. A liquid freezes
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c. AgI precipitates from a solution containing Ag+ and I- ions
17.50 Use the standard molar entropies in Appendix B to calculate So at 25oC for each
of the following reactions. Account for the sign of the entropy change in each case.
a. 2 H2O2(l)  2 H2O(1) + O2(g)
b. 2 Na(s) + Cl2(g)  2 NaCl(s)
17.58 For the vaporization of benzene, Hvap = 30.7 kJ/mol and Svap = 87.0 J/(K*mol).
Calculate Ssurr and Stotal at
a. 70oC
b. 80oC
c. 90oC
Does benzene boil at 70oC and 1 atm pressure? Calculate the normal boiling point of
benzene.
17.83 Urea, an important nitrogen fertilizer, is produced industrially by the reaction
2 NH3(g) + CO2(g) --> NH2CONH2(aq) + H2O(l)
Given that Go = -13.6 kJ, calculate G at 25oC for the following sets of
conditions:
a. 10 atm NH3, 10 atm CO2, 1.0 M urea
b. 0.10 atm NH3, 0.10 atm CO2, 1.0 M urea
Is the reaction spontaneous for the conditions in part a and/or part b?
16.19 a. How many mol NaOH are in 10.0 mL?
0.0100 L (0.100 mol/L) = 0.00100 mol NaOH in 10.0 mL
In 0.0400 L of 0.0250 M valine…
0.0400 L (0.0250 mol/L) = 0.00100 mol H2A+
there is/are (what is the proper grammar?) the equivalent number of
moles of diprotic acid. Therefore, in 10.0 mL NaOH, there is exactly
enough NaOH to reach the first equivalence point of the diprotic acid.
By definition, at the first equivalence point, the pH = (pKa1 + pKa2)/2.
Ka values given in the book…
pKa1 = -log (4.8 * 10-3) = 2.32
pKa2 = -log (2.4 * 10-10) = 9.62
pH = (2.32 + 9.62) / 2 = 5.97
b. How many mol NaOH are in 15.0 mL?
0.0150 L (0.100 mol/L) = 0.00150 mol NaOH in 15.0 mL
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To reach the next equivalence point, you’d need another equivalence of NaOH
(in other words, to reach the first equivalence point, you needed 0.00100 mol, and
to reach the second equivalence point, you’d need 0.00200 mol – double the first.)
You’re only halfway to the second equivalence point, here. But when you’re
halfway there, you’re at the half-equivalence point for the second pKa value,
meaning [HA] = [A-], and you can say that pH = pKa2.
Therefore, pH here = - log Ka2 = 9.62.
c. How many mol NaOH in 20.0 mL?
0.0200 L (0.100 mol/L) = 0.00200 mol NaOH in 20.0 mL
As mentioned in part b, this is the amount needed to reach the second
equivalence point. However, there is no easy formula to calculate pH
anymore… find [A-], and given Ka2 and Kw, determine [OH-], then
[H3O+]
You have added enough OH- to completely convert your initial acid to the
A- form:
mmol A- = (0.0250 mmol/mL)(40.0 mL) = 1.00 mmol Ais the amount that you have.
Your solution volume = 40.0 mL + 20.0 mL = 60.0 mL
Concentration of A- = 1.00 mmol/60.0 mL = 0.0167 M
Now for determining [OH-]
A- + H2O  
HA + OH----------------------------------------------------------------------I(M)
0.0167
0
~0
C(M)
-x
+x
+x
E(M)
0.0167 – x
x
x
Kb = Kw/Ka2 = 1.0 * 10-14/2.4 * 10-10 = 4.17 * 10-5
and Kb = [HA][OH-]/[A-] = 4.17 * 10-5 = x2/0.0167 – x
Assume x is small… x2 = 6.96 * 10-7
x = 8.34 * 10-4
[OH-] = 8.34 * 10-4 M
[H3O+] = Kw/[OH-] = 1.0 * 10-14/8.34 * 10-4 = 1.2 * 10-11 M
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and pH = 10.92
16.92 Ag2CO3 dissociates into 2 Ag+ and 1 CO32So, [Ag=] = 2.56 * 10-4 M = 2 * [CO32-]
[CO32-] = 1.28 * 10-4 M
and Ksp = [Ag+] 2[CO32-] = (2.56 * 10-4)2(1.28 * 10-4) = 8.39 * 10-12
16.96 a. Provides a source of Ag+ and shifts the equilibrium to the left, therefore
decreases solubility.
b. Provides a source of acid, which will tie up the CO32- ions in the form of HCO3and remove them from the equilibrium, and shift the equilibrium to the right, increasing
solubility.
c. Provides a source of carbonate ions and shifts the equilibrium to the left,
therefore decreases solubility.
d. Adding NH3 will form a complex ion with Ag+ ions and drives the reaction to
the right, increasing solubility.
16.104
AgI(s)   Ag+(aq) + I-(aq)
Ksp = 8.5 * 10-17
Ag+(aq) + 2 CN-(aq)  Ag(CN)2-(aq)
Kf = 3.0 * 1020
-------------------------------------------------------AgI(s) + 2 CN-(aq)   Ag(CN)2-(aq) + I-(aq)
a.
K for this reaction = (Ksp)(Kf) = 2.6 * 104
16.108 Determine the concentration of each of the ions, then determine the IP, and
compare to Ksp (which, for BaSO4, is 1.1 * 10-10)
Your total volume is 400 mL.
Mol Ba2+ = 0.100 L(4.0 * 10-3 mol/L) = 4.0 * 10-4 mol
Mol SO42- = 0.300 L(6.0 * 10-4 mol/L) = 1.8 * 10-3 mol
[Ba2+] = 4.0 * 10-4 mol/0.4 L = 1.0 * 10-3 M
[SO42-]= 1.8 * 10-3 mol/0.4 L = 4.5 * 10-4 M
IP = [Ba2+][SO42-] = (1.0 * 10-3)(4.5 * 10-4) = 4.5 * 10-7
Since IP > Ksp a precipitate will form.
16.109 Remember that the concentrations of each of the ions are cut in half because
you’re mixing an equal amount of each solution.
Ksp for PbCl2 = 1.2 * 10-5 = [Pb2+][Cl-] 2
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and IP = (0.005)(0.005)2 = 1.2 * 10-7
This IP is less than Ksp, so no precipitate will form.
To figure out the minimum amount of Cl- needed for precipitation… use Ksp in
the equation and solve for [Cl-]
[Cl-] = (Ksp/[Pb2+])1/2 = 0.049 M
If you add Cl- to 0.049 M or greater, a precipitate will form.
16.112 First, use the pH to figure out the concentration of [OH-] present.
pH = -log [H3O+]
[H3O+] = 10-10.80 = 1.6 * 10-11 M
[OH-] = Kw/[H3O+] = 6.2 * 10-4 M
Ksp of Mg(OH)2 = 5.6 * 10-12
At pH = 10.80, IP = [Mg2+][OH-] 2 = (2.5 * 10-4)(6.2 * 10-4)2 = 9.6 * 10-11
And since IP > Ksp , at pH = 10.80, a precipitate will form.
16.131 Strategy: figure out how much Mg2+ is dissolved, since 1 mol Mg2+ exists for
each mol Mg(OH)2.
You’re given the pH, so you can determine the concentration of OH-.
pH = 9.00 [H3O+] = 10-9.00 = 1.0 * 10-9 M
[OH-] = Kw/[H3O+] = 1.0 * 10-5 M
Ksp = [Mg2+][OH-] 2 = x(1.0 * 10-5)2 = 5.6 * 10-12 (value for Ksp would be given)
Solve for x, which is the molar solubility for the compound…
x = 0.056 M
17.8 a. Assuming standard conditions for pressure, and we are given a designated
temperature, G = H – TS = 58,500 J/mol – (598 K)(92.9 J/K*mol) = 2900 J/mol
Yes – boiling is spontaneous.
b. Boiling point = equilibrium point where G = 0
0 = H – TS
H = TS
T = H/S = 58.500 J/mol/92.9 J/(K*mol) = 357oC
17.38
a. Change from solid to gas… positive.
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b. Change from liquid to solid… negative.
c. Change from aqueous ions to solid… negative.
17.50 a. So = [2 So(H2O(l)) + So(O2)] – 2 So(H2O2)
= [2 mol (69.9 J/K*mol) + (1 mol)(205.0 J/K*mol)] – (2 mol)(110
J/K*mol) = +125 J/K*mol
The change in entropy is positive, which makes sense because you’re
forming a gas from a liquid.
b. So = 2 So (NaCl) – [2 So(Na) + So(Cl2)]
= (2 mol)(72.1 J/K*mol) – [(2 mol)(51.2 J/K*mol) + (1 mol)(223.0
J/K*mol)]
=-181.2 J/K*mol
The change in entropy is negative, which makes sense, since a gas is being
turned into a solid.
17.58 In each case, Ssurr = -Hvap/T and Stotal = Svap + Ssurr
a. Ssurr = (-30,700 J/mol)/343 K = -89.5 J/(K*mol)
Stotal = 87.0 J/(K*mol) + (-89.5 J/K*mol) = -2.5 J/K*mol
b. Ssurr = (-30,700 J/mol)/353 K = -87.0 J/K*mol
Stotal = 87.0 J/K*mol + (-87.0 J/K*mol) = 0
c. Stotal = (-30,700 J/mol)/363 K = -84.6 J/K*mol
Stotal = 87.0 J/K*mol + (-84.6 J/K*mol) = +2.4 J/K*mol
Boiling will not be spontaneous at 70oC because Stotal is negative. The normal
boiling point = equilibrium point = 80oC.
17.83 G = Go + RT ln ([urea]/PNH32PCO2)

a. = -13.6 kJ/mol + (8.314 * 10-3 kJ/K*mol)(298 K) ln (1.0)/(10)2(10) = -30.7
kJ/mol
b. = -13.6 kJ/mol + (8.314 * 10-3 kJ/K*mol)(298 K) ln (1.0)/(0.10)2(0.10) = +3.5
kJ/mol
Case a: spontaneous. Case b: nonspontaneous. You can tell by the sign of G
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