Department of Chemistry, Geosciences and Environmental Sciences

 

                                                                          Exam 2-A

                       

Chemisty 1084:  Section 010                                                                                  Spring 2007

 

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.  The conjugate base of HSO4¯ is:

       (a)  OH¯                (b)  H2SO4             (c)  SO42-              (d)  HSO4+             (e)  H3SO4+

 

Answer:  C

 

2.  What is the hydrogen ion concentration in a solution with a pH = 5.64?

       (a)  5.64 M                                                                      (c)  2.3 × 10-6 M

       (b)  4.3 × 10-9 M                                                              (d)  8.36 M

                                     (e)  1.0 × 10-7 M

 

Answer:  C

 

 

3.  Which of the following is the weakest acid?

       (a)  Pyruvic acid (pKa = 4.25)                                         (cHeptanoic acid (pKa = 4.89)

       (b)  Adipamic acid (pKa = 4.63)                                     (d)  Acetic acid (pKa = 4.76)

 

Answer:  C

The acid with the largest value of pKa is the weakest acid.

 

SO2Cl2(g)    SO2(g)  +  Cl2(g)          DH = 67 kJ/mol

Questions 4 and 5 pertain to a reaction mixture of the reaction shown above that is in equilibrium:

 

4.  An increase in pressure will ___________________________.

       (a)  shift the reaction mixture towards the products.

       (b)  shift the reaction mixture towards the reactants.

       (c)  have no effect on the equilibrium mixture.

 

Answer:  B

 

An increase in pressure favors the side with the fewer number of moles of gas which is the reactant side.

 

5.  An increase in temperature will ___________________________.

       (a)  shift the reaction mixture towards the products.

       (b)  shift the reaction mixture towards the reactants.

       (c)  have no effect on the equilibrium mixture.

 

Answer:  A

 

An increase in temperature favors the endothermic direction which is the forward direction.

 

6.  Which of the following salts will dissolve in water to produce a basic solution?

       (a)  NH4NO3                                                                    (cNaCl

       (b)  KBr                                                                           (dLiF

 

Answer:  D

 

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

       (a)  equal to 7.0                                                               (c)  below 7.0

       (b)  above 7.0       

 

Answer:  C

 

This is a weak base-strong acid titration.

 

8.  Which of the following salts will have the lowest solubility?

       (a)  BaCO3  (Ksp = 5.0 × 10-9)                                          (c)  MgCO3 (Ksp = 3.5 × 10-8)

       (b)  CaCO3  (Ksp = 4.5 × 10-9)                                         (d)  PbCO3  (Ksp = 7.0 × 10-14)

                                     (e)  NiCO3  (Ksp = 1.3 × 10-7)

 

Answer:  D

 

The salt with the smallest value of Ksp.

 

9.  The Earth’s atmosphere is split into layers based on:

(a)  the way that pressure and temperature fluctuate with altitude.

(b)  the way that temperature fluctuates with altitude.

(c)  the way that pressure fluctuates with altitude.

(d)  altitude only.

 

Answer:  B

 

10.  Of the following, only ______________ does not result in the formation of acid rain.

       (a)  carbon dioxide                                                          (c)  sulfur dioxide

       (b)  nitrogen monoxide                                                    (d)  nitrogen dioxide

                                     (e)  methane

 

Answer:  E

 

 

 


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 a 0.045 M Ba(OH)2 solution?

 

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

[OH¯] = 2 × [Ba(OH)2] = 2 × 0.045 M = 0.090 M

pOH = –log(0.090) = 1.05

pH = 14.00 – 1.05 = 12.95 (2 digits past the decimal)


12.  (15 points)  At 500 K, Kc = 0.61 for the reaction:

PCl5(g)   PCl3(g)  +  Cl2(g)

       Initially, a 1.50 L flask is filled with 0.400 mol of PCl5(g) and allowed to equilibrate at 500 K.  What are the equilibrium concentrations of PCl5, PCl3, and Cl2?

 

Calculate initial concentration of PCl5

Setup the equilibrium problem:

 

PCl5(g)

PCl3(g)

+

Cl2(g)

initial

0.267

 

0

 

0

change

x

 

+x

 

+x

equilibrium

0.267 – x

 

x

 

x

 

Let x = D[PCl3].  Calculate the other changes in concentration.  Add the initial and change in concentration to obtain equilibrium concentrations.  Place the equilibrium concentrations into the equilibrium expression and solve for x:

Throw out the negative answer.  The equilibrium concentrations are: 

[PCl5] = 0.267 – 0.20 = 0.07 M (0.067 would also be accepted)

[PCl3] = 0.20 M

[Cl2] = 0.20 M

 


13.  (10 points)  What is the pH of a solution that is 0.15 M in HCN and 0.25 M in NaCN.  The Ka for HCN is equal to 4.9 × 10-10.

 

Common ion approach:

Set up the weak acid problem:

 

HCN(aq)

H+(aq)

+

CN¯(aq)

initial

0.15

 

0

 

0.25

change

x

 

+x

 

+x

equilibrium

0.15–x

 

x

 

0.25 + x

 

Let x = D[H+].  Calculate the other changes in concentration.  Add the initial and change in concentration to obtain equilibrium concentrations.  Place the equilibrium concentrations into the equilibrium expression and solve for x:

Calculate the pH:  pH = –log(2.9×10-10) = 9.53  (2 digits past the decimal point)

-OR-

This problem may be solved using the Henderson-Hasselbach equation: 

pKa = –log(4.9×10-10) = 9.31

[base] = [NaCN] = 0.25 M

[acid] = [HCN] = 0.15 M

 

 

 

14.  (10 points)  What is the solubility, in grams per liter, of cobalt(II) phosphate, Co3(PO4)2.  The Ksp for cobalt(II) phosphate is equal to 2.05 × 10-35.

 

Setting up the solubility equilibrium reaction:

 

Co3(PO4)2(s)

3 Co2+(aq)

+

2 PO43-(aq)

initial

 

0

 

0

change

 

+3x

 

+2x

equilibrium

 

3x

 

2x

 

Let x be equal to the molar solubility of Co3(PO4)2.  Calculate the other changes in concentration.  Add the initial and change in concentration to obtain equilibrium concentrations.  Place the equilibrium concentrations into the equilibrium expression and solve for x:

4.53 × 10-8 mol/L of Co3(PO4)2 dissolve in water.  To convert to grams per liter, multiply by the molar mass of Co3(PO4)2:

(4.53 × 10-8 mol/L)(366.7 g/mol) =1.66 × 10-5 g/L


15.  (15 points)  What is the pH of a 0.25 M KF solution?  The Ka for HF is equal to 6.8 × 10-4.

 

KF is a strong electrolyte:  KF(aq) ¾¾®  K+(aq)  +  F¯(aq)

 

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

Setting up the weak base equilibrium:

 

(aq)

+

H2O(l)

HF(aq)

+

CN¯(aq)

initial

0.25

 

 

0

 

0

change

x

 

 

 

+x

 

+x

equilibrium

0.25–x

 

 

 

x

 

x

Let x = D[OH¯].  .  Calculate the other changes in concentration.  Add the initial and change in concentration to obtain equilibrium concentrations.  Place the equilibrium concentrations into the equilibrium expression and solve for x:

Calculate the pH:  pOH = –log(1.9 × 10-6) = 5.71

pH = 14.00 – 5.71 = 8.29 (2 digits past the decimal point)


(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.

 

 

LeChatelier’s Principle:  a reaction in equilibrium will respond to a disturbance in such a fashion such that the disturbance is partially counteracted.

 

 

 

 

 

Lewis base:  a electron pair donor.

 

 

 

 

Endpoint of a titration:  the point in a titration where the pH indicator changes color.