Department of Chemistry, Geosciences and Environmental Sciences

 

                                                                     Exam 2-A-Key

                       

Chemisty 1084:  Section 010                                                                                       Fall 2006

 

Name:________________________________________________________

 

Read all directions and questions carefully!!  This exam consists of two parts.  The first part consists of 10 multiple choice questions worth four points each for a total of 40 points.  The second part consists of five numerical problems worth either 10 or 15 points per question for a total of 60 points.  Show all your work necessary for the numerical problems as partial credit will be given for those problems.

 

Possibly Useful Constants and Equations

 

            Kw = [H+][OH¯] = 1.00 × 10-14 at 25°C

            For an equation of the form:  ax2 + bx + c = 0: 

            Henderson-Hasselbach equation: 

                   

Score

 

                                    Part 1 (40 points):_____________________

 

 

                                    Part 2 (60 points):_____________________

 

 

                                    Total (100 points):_____________________

 

 

Don’t forget to put your name on this test!

 

Good Luck!!

Part 1

 

Multiple Choice

 

Please indicate the answer to each question by putting your choice in the space provided.  There is only one correct answer for each question.  There will be 10 multiple choice questions worth 4 points each.

 

1.  A Brønsted-Lowry base is defined as a substance that ___________________.

       (a)  increases [H+] when dissolved in water.

       (b)  increases [OH¯] when dissolved in water.

       (c)  acts as a proton acceptor.

       (d)  acts as a proton donor.

 

Answer:  C

definition of a Bronsted base.

 

2.  Which one of the following is not a strong acid?

       (a)  HCl                                                                      (c)  HClO4

       (b)  HNO2                                                                   (d)  HBr

                              (e)  All of the above are strong acids.

 

Answer:  B

 

3.  Which one of the following is the strongest acid?

       (a)  HF (pKa = 3.17)                                                   (c)  HNO2 (pKa = 3.35)

       (b)  HCN  (pKa = 9.31)                                              (d)  HC2H3O2  (pKa= 4.74)

                              (e)  All of these acids have equal strengths.

 

Answer:  A

The acid with the lowest pKa value is the strongest.

 

4.  Which one of the following pairs of compounds will produce a buffer solution when dissolved in water?

       (a)  HBr, LiBr                                                             (c)  HF, KF

       (b)  HI, CsI                                                                 (d)  HCl, RbCl

                  (e)  All of the combinations above produce a buffer solution

 

Answer:  C

A buffer consists of a weak acid and its conjugate base.  The only weak acid above is HF.

 

5.  The pH at the equivalence point in the titration of 25.00 mL of 0.125 M HCHO2  (Ka = 1.8 × 10-4) versus 0.100 M NaOH will be ___________________.

       (a)  equal to 7.00.                                                       (c)  below 7.00

       (b)  above 7.00.                                                          (d)  impossible to determine without more data.

 

Answer:  B

This is a weak acid-strong base titration.  The pH at the equivalence point is above 7.

6.  Which of the following salts has the largest molar solubility?

       (a)  MgCO3  (Ksp = 3.5 × 10-8)                                    (c)  CaCO3  (Ksp = 4.5 × 10-9)

       (b)  BaCO3  (Ksp = 5.0 × 10-9)                                    (d)  CuCO3  (Ksp = 2.3 × 10-10)

 

Answer:  A

The salt in this series with the largest Ksp value has the largest molar solubility.

 

7.  Consider the following reaction in equilibrium:

2 CO2(g)    2 CO(g) +  O2(g)    DH = –514 kJ

       Increasing the pressure on the equilibrium mixture will ________________________.

       (a)  shift the reaction towards products.

       (b)  shift the reaction towards reactants.

       (c)  have no effect on the reaction mixture.

 

Answer:  B

Increasing the pressure favors the side with the fewer number of moles of gas.  In this case, it is the reactants.

 

8.  Rainwater is naturally acidic mainly due to the presence of ____________ in the atmosphere.

       (a)  N2                                                                         (c)  O2

       (b)  O3                                                                         (d)  CO2

                              (e)  CF2Cl2

 

Answer:  D

From Chapter 18.  Will be covered on Exam 3 in Fall 2007.

 

9.  Photochemical smog is the result of the emission of which gas from internal combustion engines?

       (a)  SO2                                                                       (c)  NO

       (b)  CO2                                                                      (d)  O3

                              (e)  CO

 

Answer:  C

From Chapter 18.  Will be covered on Exam 3 in Fall 2007

 

10.  The quality of freshwater is measured by the amount of which dissolved gas in the water?

       (a)  O3                                                                         (c)  CO2

       (b)  N2                                                                         (d)  O2

                              (e)  NH3

 

Answer:  D

From Chapter 18.  Will be covered on Exam 3 in Fall 2007.

 

 

 

 

Part 2

 

Numerical Problems

 

Solve the following problems, keeping track of significant figures where applicable.  Please show all the work necessary to obtain your answer in order to receive partial credit for possibly wrong answers.  Generally, full credit will not be given for the correct answer without any of the work performed to obtain the answer being shown on the paper.  Each question is worth either 10 or 15 points.

 

11.  (10 points)  What is the pH of 0.0015 M Ba(OH)2?

 

Strong base problem:  Ba(OH)2(aq)  ¾¾®  Ba2+(aq)  +  2 OH¯(aq) 

[OH¯] = 2 × [Ba(OH)2] = 2 × 0.0015 M = 0.0030 M

pOH = – log(0.0030) = 2.52

pH = 14.00 – 2.52 = 11.48 (2 digits past decimal)

12.  (15 points)  (a)  (9) Calculate the solubility, in grams per liter, of CaF2 in pure water.  The Ksp for CaF2 is equal to 3.9 × 10-11.

 

Setting up the solubility equilibrium:

 

CaF2(s)

Ca2+(aq)

+

2 F¯(aq)

initial concentration

–

 

0

 

0

change in concentration

–

 

+ x

 

+ 2x

equilibrium concentration

–

 

x

 

2x

 

Let x = mol/L of CaF2 that dissolve, then D[Ca2+] = +x and D[F¯] = +2x.  Add the initial and change in concentrations to obtain equilibrium concentrations.  Place the equilibrium concentrations in the Ksp expression and solve for x:

2.1 × 10-4 mol/L of CaF2 dissolve.  Convert this to grams per liter by multiplying by the molar mass of CaF2:

 

(b)  (6 points)   Calculate the solubility, in grams per liter, of CaF2 in a 0.10 M Ca(NO3)2 solution .  The Ksp for CaF2 is equal to 3.9 × 10-11.

 

This is a common ion problem for solubility.  The initial concentration of Ca2+ will be equal to 0.10 M from the 0.10 M Ca(NO3)2 solution.  Setting up the equilibrium problem:

 

CaF2(s)

Ca2+(aq)

+

2 F¯(aq)

initial concentration

–

 

0.10

 

0

change in concentration

–

 

+ x

 

+ 2x

equilibrium concentration

–

 

0.10 + x

 

2x

 

Let x = mol/L of CaF2 that dissolve, then D[Ca2+] = +x and D[F¯] = +2x.  Add the initial and change in concentrations to obtain equilibrium concentrations.  Place the equilibrium concentrations in the Ksp expression and solve for x:

9.9 × 10-6 mol/L of CaF2 dissolve.  Convert this to grams per liter by multiplying by the molar mass of CaF2:

 

13.  (10 points)  Calculate the pH of a buffer solution that is 0.15 M in HC2H3O2 and 0.20 M in NaC2H3O2.  The Ka for HC2H3O2 is equal to 1.8 × 10-5.

This problem can be solved via a common ion approach.  Set up the weak acid equilibrium with initial concentrations of HC2H3O2­ and C2H3O2¯ equal to 0.15 M and 0.20 M respectively.

 

HC2H3O2(aq)

H+(aq)

+

C2H3O2¯(aq)

initial concentration

0.15

 

0

 

0.20

change in concentration

–x

 

+x

 

+x

equilibrium concentration

0.15 – x

 

x

 

0.20 + x

 

Let x = D[H+].  Then the other changes in concentration can be determined.  Add the initial plus the change in concentration to obtain equilibrium concentrations.  Place the equilibrium concentrations in the Ka expression and solve for x:

Calculate the pH of the solution:  pH = –log[H+] = –log(1.4×10-5) = 4.87 (2 decimal places)

 

The other method to do this problem is use the Henderson-Hasselbach equation:

pKa = –log(1.8×10-5) = 4.74

[base] = [C2H3O2¯] = 0.20 M

[acid] = [HC2H3O2] = 0.15 M

 

14.  (10  points)  Calculate both the [H+] and [OH¯] at a pH of 8.45.

 

Calculate [H+] from the pH:

Calculate [OH¯] from [H+] and Kw:

15.  (15 points)  Calculate the pH of a 0.025 M HCNO solution.  The Ka for HCNO is equal to 3.5 ×

10-4.

 

This is a weak acid problem.  Setting up the weak acid equilibrium:

 

HNO2(aq)

H+(aq)

+

NO2¯(aq)

initial concentration

0.025

 

0

 

0

change in concentration

–x

 

+x

 

+x

equilibrium concentration

0.025 – x

 

x

 

x

Let x = D[H+].  Then the other changes in concentration can be determined.  Add the initial plus the change in concentration to obtain equilibrium concentrations.  Place the equilibrium concentrations in the Ka expression and solve for x:

 

The negative answer gets thrown out.  Therefore, [H+] = 2.8 × 10-3 M.  The pH is equal to:

pH = –log(2.8×10-3) = 2.55

 

(10 points extra credit)  Define 2 out the 3 following terms.  Credit will be given only for the complete definition.  No partial credit will be given.  Choose only 2 out of the three terms.  If you give definitions for all three terms without clearly noting which ones you want graded, only the first two terms will be graded.

 

 

Common ion effect:  the dissociation of a weak acid or weak base being suppressed by the presence of a strong electrolyte that has a ion in common with the acid or base.

 

 

 

 

Le Châtelier’s Principle:  a reaction in equilibrium will respond to a disturbance in such a way so that the disturbance is partially counteracted.

 

 

 

 

 

Lewis Acid:  an electron pair donor.