Department of Chemistry, Geosciences and Environmental Sciences

 

                                                                     Exam 2-A-Key

                       

Chemisty 1084:  Section 010                                                                                  Spring 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

            Quadratic Formula:  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.  Which of the following choices is the definition of an acidic aqueous solution?

       (a)  [H+][OH¯] = Kw                                                   (c)  [H+] < [OH¯]

       (b)  [H+] = [OH¯]                                                       (d)  [H+] > [OH¯]

                                                            (e)  [OH¯] = 0

 

Answer:  C

 

2.  A solution with a pH equal to ___________ has the largest concentration of hydroxide ions.

       (a)  3.21                                                                      (c)  7.93

       (b)  12.62                                                                    (d)  9.82

                                                            (e)  7.00

 

Answer:  B

The solution with the largest pH has the highest hydroxide ion concentration.

 

Questions 3 and 4 refer to the following reaction at equilibrium:

2 SO2(g)  +  O2(g)    2 SO3(g)       DH = –99 kJ

 

3.  An increase in pressure will:

       (a)  cause the reaction to move towards products.

       (b)  cause the reaction to move towards reactants.

       (c)  have no effect.

 

Answer:  A

An increase in pressure will favor the side with the smaller number of moles of gas.  This is the product side.

 

4.  A decrease in temperature will:

       (a)  cause the reaction to move towards products.

       (b)  cause the reaction to move towards reactants.

       (c)  have no effect.

 

Answer:  A

A decrease in temperature will favor the exothermic direction.  Since DH for this reaction is negative, this is the forward direction towards products.

 

5.  Which of the following acids is the weakest acid?

       (a)  HF  (Ka = 6.8 × 10-4)                                            (c)  HNO2  (Ka = 4.5 × 10-4)

       (b)  HClO  (Ka = 3.0 × 10-8)                                       (d)  HCN  (Ka = 4.9 × 10-10)

                                                            (e)  HC2H3O2  (Ka = 1.8 × 10-5)

 

Answer:  D

The weakest acid has the smallest Ka value.

 

6.  Which of the following salt solutions will be basic?

       (a)  AlCl3                                                                    (c)  KNO3

       (b)  NH4Cl                                                                  (d)  CsF

 

Answer:  D

AlCl3 and NH4Cl contain an acidic cations.  They will be acidic.  KNO3 will give a neutral solution.  CsF has a basic anion.

 

7.  Which of the following could be added to a solution of sodium acetate, NaC2H3O2, to produce a buffer solution?

                        HC2H3O2                     HCl                       KC2H3O2                           NaCl

       (a)  HC2H3O2 only                                                      (b)  HC2H3O2 or HCl

       (c)  HCl only                                                              (d)  KC2H3O2 only

                                                            (e)  NaCl or KC2H3O2

 

Answer:  A

A buffer solution contains either a weak acid and its conjugate base or a weak base and its conjugate acid.  NaC2H3O2 contains C2H3O2¯ ion which is the conjugate base of the weak acid of HC2H3O2.

 

8.  The pH of the equivalence point in the titration of 25.00 mL of 0.125 M HCl with 0.150 M HONH2  (Kb = 1.1 × 10-8) will be:

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

       (b)  above 7.00.

 

Answer:  C

The pH at the equivalence point in a strong acid-weak base titration is below 7.0.

 

9.  Select the substance that is thought to be partially responsible for depleting the concentration of ozone in the stratosphere:

       (a)  CFCl3                                                                   (c)  O2

       (b)  CO2                                                                      (d)  N2

                                                            (e)  He

 

Answer:  A

 

10.  Which one of the following pollutants is produced from incomplete combustion of hydrocarbons?

       (a)  CO2                                                                      (c)  SO2

       (b)  NO                                                                       (d)  CO

                                                            (e)  O3

 

Answer:  D

 

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 0.0026 M Ca(OH)2 solution?

 

This is a strong base.  Ionization reaction:  Ca(OH)2(aq)  ¾®  Ca2+(aq)  +  2 OH¯(aq)

 

Calculate [OH¯]:  [OH¯] = 2 × [Ca(OH)2] = 2 × 0.0026 M = 0.0052 M

 

Calculate pOH :  pOH = –log(0.0052) = 2.28 (2 digits past the decimal point)

 

12.  (15 points)  What is the pH of a 0.15 M KF solution?  The Ka for HF is equal to 6.8 × 10-4.

 

KF is a salt that dissolves in water to produce ions:  KF(aq)  ¾¾®  K+(aq)  +  F¯(aq)

 

F¯ is the conjugate base of HF.  It will have a Kb value equal to: 

Calculating the pH of 0.15 M F¯ will be a weak base equilibrium problem:

 

 

F¯(aq)

+

H2O(l)

 

HF(aq)

+

OH¯(aq)

initial concentration

0.15

 

–

 

0

 

0

change in concentration

–x

 

–

 

+x

 

+x

equilibrium concentration

0.15 – x

 

–

 

x

 

x

 

Let x = D[OH¯].  Then, calculate the changes in concentrations of HF and F¯.  Add the initial concentrations and the change in concentrations to obtain the equilibrium concentrations in terms of x.  Place these concentrations into the Kb expression and solve for x:

Take the negative log of this number to calculate pOH:  pOH = –log(1.5 × 10-6) = 5.82

Subtract the pOH from 14.00 to obtain the pH:  pH = 14.00 – 5.82 = 8.18  (2 digits past the decimal point)

13.  (10 points)  What is the pH of a 0.10 M (CH3)2NH solution.  The Kb for (CH3)2NH is equal to 5.4 × 10-4.

 

This is a weak base problem because you are given a “Kb” value.  Setting up the problem:

 

(CH3)2NH(aq)

+

H2O(l)

 

(CH3)2NH2+(aq)

+

OH¯(aq)

initial concentration

0.10

 

–

 

0

 

0

change in concentration

–x

 

–

 

+x

 

+x

equilibrium concentration

0.10 – x

 

–

 

x

 

x

 

Let x = D[OH¯].  Then, calculate the changes in concentrations of (CH3)2NH2+ and (CH3)2NH.  Add the initial concentrations and the change in concentrations to obtain the equilibrium concentrations in terms of x.  Place these concentrations into the Kb expression and solve for x:

Take the negative log of this number to calculate pOH:  pOH = –log(7.3 × 10-3) = 2.13

Subtract the pOH from 14.00 to obtain the pH:  pH = 14.00 – 2.13 = 10.87  (2 digits past the decimal point)

 

 

14.  (10 points)  What is the pH of a solution that is 0.10 M in HC2H3O2 and 0.15 M in KC2H3O2?  The Ka for HC2H3O2 is 1.8 × 10-5.

This problem can be solved by two methods.  It is a common ion problem and a buffer solution.  First, performing a common ion problem:

Initial concentrations:  [HC2H3O2] = 0.10 M and [C2H3O2¯] = 0.15 M (from KC2H3O2 which is a salt)

 

HC2H3O2(aq)

H+(aq)

+

C2H3O2¯(aq)

initial concentration

0.10

 

0

 

0.15

change in concentration

–x

 

+x

 

+x

equilibrium concentration

0.10 – x

 

x

 

0.15 + x

 

Let x =D[H+].  Calculate the change in concentrations of C2H3O2¯ and HC2H3O2.  Add the initial and change in concentrations to calculate equilibrium concentrations in terms of x.  Place the equilibrium concentrations into the Ka expression and solve for x:

Calculate the pH of the solution:  pH = –log(1.2 × 10-5) = 4.92 (2 digits past the decimal place)

 

Since this is also a buffer solution, the pH could also be calculated using the Henderson-Hasselbach equation:

pKa = –log(1.8 × 10-5) = 4.74

[base] = [C2H3O2¯] = 0.15 M

[acid] = [HC2H3O2] = 0.10 M

 

 

15.  (15 points)  What is the solubility, in grams per liter, of BaF2 in (a) pure water, and (b) in 0.100 M BaCl2 solution.  The Ksp for BaF2 is equal to 1.84 × 10-7.

(a)  Setting up the solubility equilibrium problem:

 

BaF2(a)

Ba2+(aq)

+

2 F¯(aq)

initial concentration

–

 

0

 

0

change in concentration

–

 

+x

 

+2x

equilibrium concentration

–

 

x

 

 2x

 

Let x be equal to the mole per liter of BaF2 that dissolve.  Then [Ba2+] increases by x and [F¯] increases by 2x (because there are 2 F¯ ions in BaF2).  Calculate the equilibrium concentrations and put them in the Ksp expression and solve for x:

Therefore, 3.58 × 10-3 moles per liter of BaF2 dissolve.  Multiply this by the molar mass of BaF2 to calculate the grams per liter of BaF2 that dissolve:

(3.58 × 10-3 mol/L)(175.3 g/mol) = 0.628 g/L  (3 sig figs)

 

(b)  This is a common ion problem with sparingly soluble salts.  The initial concentration of Ba2+ will be equal to 0.100 M (from BaCl2).  Setting up the solubility equilibrium problem:

 

BaF2(a)

Ba2+(aq)

+

2 F¯(aq)

initial concentration

–

 

0.10

 

0

change in concentration

–

 

+x

 

+2x

equilibrium concentration

–

 

0.10 + x

 

 2x

 

Let x be equal to the mole per liter of BaF2 that dissolve.  Then [Ba2+] increases by x and [F¯] increases by 2x (because there are 2 F¯ ions in BaF2).  Calculate the equilibrium concentrations and put them in the Ksp expression and solve for x:

Therefore, 6.78 × 10-4 moles per liter of BaF2 dissolve.  Multiply this by the molar mass of BaF2 to calculate the grams per liter of BaF2 that dissolve:

(6.78 × 10-4 mol/L)(175.3 g/mol) = 0.119 g/L  (3 sig figs)

 

 

(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.

 

Le Châtelier’s Principle:  A reaction in equilibrium will respond to a disturbance in such a way as to partially counteract the disturbance.

 

 

 

 

 

Buffer capacity:  The minimum amount of acid or base required to overwhelm a buffer solution.

 

 

 

 

 

Lewis Acid:  An electron pair acceptor or an electrophile.