Department of Chemistry, Geosciences and Environmental Sciences

 

                                                                     Exam 2-A-Key

                       

Chemisty 1084:  Section 010                                                                                         Fall 2009

 

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 value of Kp is 7.0 at 400 K for the reaction: 

Br2(g)  +  Cl2(g) ƒ  2 BrCl(g)

       A closed vessel is filled with 1.0 atm of Br2, 1.0 atm of Cl2, and 3.0 atm of BrCl at 400 K.  Use  Qp to determine which of the following statements is true.

(a)  The reaction will move in the forward direction to reach equilibrium.

(b)  The reaction will move in the reverse direction to reach equilibrium.

(c)  The reaction mixture is already at equilibrium.

 

Answer:  B  (because the value of Q = 9.0 which is larger than the value of Kp)

 

2.  The reaction below is endothermic:

N2(g)  + 2 O2(g)  ƒ  2 NO2(g)

       For an equilibrium mixture, an increase in temperature will:

       (a)  move the reaction towards products.

       (b)  move the reaction towards reactants.

       (c)  have no effect.

 

Answer:  A  (an increase in temperature favors the endothermic direction)

 

3.  For the equilibrium mixture described in the previous question, a decrease in pressure will:

       (a)  move the reaction towards products.

       (b)  move the reaction towards reactants.

       (c)  have no effect.

 

Answer:  B  (a decrease in pressure favors the side with the greater number of moles of gas)

 

4.  The conjugate acid of HSO4¯ is:

       (a)  SO42-                                                                          (c)  HSO4+

       (b)  H2SO4                                                                       (d)  H+

                                     (e)  HSO3+

 

Answer:  B  (add an H+ to HSO4¯)

 

5.  An acidic solution has:

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

       (b)  [H+] = 0                                                                     (d)  [H+] > [OH¯]

 

Answer:  D

 

 

6.  What is the concentration of hydrogen ions in a solution with a pH = 9.777?

       (a)  5.98 ×10-3                                                                  (c)  1.67 ×104

       (b)  1.67 × 10-10                                                               (d)  1.00 ×10-7

                                     (e)  9.78 × 10-14

 

Answer:  B  (10-9.777 = 1.67 × 10-10)          

 

7.  A 0.1 M solution of ____________  will be acidic.

       (a)  Na2S                                                                          (c)  KCl

       (b)  NaNO3                                                                      (d)  NH4Cl

 

Answer:  D  (NH4+ is an acidic cation, it is the conjugate acid of the weak base NH3)

 

 

8.  Which of the following is the strongest acid?

       (a)  HC2H3O2  (pKa = 4.74)                                             (c)  HClO  (pKa = 7.52)

       (b)  HF (pKa = 3.17)                                                        (d)  HCHO2  (pKb = 3.74)

 

Answer:  B  (Smallest pKa value corresponds to strongest acid)

 

9.  Which one of the following pairs cannot be mixed together to form a buffer solution?

       (a)  NH3 and NH4Cl                                                        (c)  HC2H3O2 and NaC2H3O2

       (b)  H3PO4 and KH2PO4                                                 (d)  KBr and HBr

                        (e)  All of the combinations above will form a buffer.

 

Answer:  D  (HBr is a strong acid.  Strong acids do not form buffers)

 

10.  The pH at the equivalence point for a titration of a weak base with a strong acid will be:

       (a)  equal to 7.0                                                               (c)  below 7.0

       (b)  above 7.0

 

Answer:  C

 

 

 

 

 

 

 

 

 

 

 

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)  The value of Kp is 0.78 at 400 K for the reaction:

NH4Br(s)  ƒ  NH3(g)  +  HBr(g)

       Initially, a large amount of NH4Br is placed in an evacuated flask and allowed to equilibrate at 400 K.  What are the equilibrium partial pressures of NH3 and HBr?

 

Setup the equilibrium problem:

 

NH4Br(s)

ƒ

NH3(g) +

HBr(g)

Initial

–

 

0

0

Change

–

 

+x

+x

Equilibrium

–

 

x

x

Since NH4Br is a solid, it does not appear in the equilibrium expression.  Let x = DPNH3, then x =DPHBr.  Add the change and initial pressures to get equilibrium pressures.  Place the equilibrium pressures into the equilibrium expression and solve for x:

Therefore, at equilibrium:  PNH3 = PHBr = 0.88 atm  (2 sig figs)

 

12.  (15 points)  What is the pH of a 0.025 M solution of nitrous acid (HNO2).  The Ka for nitrous acid is equal to 4.5 × 10-4.

 

Weak acid problem.  Setting up the equilibrium:

 

 

HNO2(aq)

ƒ

H+(aq) +

NO2¯(aq)

Initial

0.025

 

0

0

Change

–x

 

+x

+x

Equilibrium

0.025 – x

 

x

x

 

Let x = D[H+].  Calculate all the other changes and then the equilibrium concentrations.  Place the equilibrium concentrations into the Ka expression and solve for x:

You cannot neglect the “–x” in this problem!!

 

 

 

 

 

x = 3.0×10-3  or x = –3.5 × 10-3

Throwing out the negative answer:

x = 3.0 × 10-3 = [H+]

pH = –log(3.0×10-3) = 2.52 (2 digits past decimal)

        

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

13.  (10 points)  What is the pH of a 0.055 M Ba(OH)2 solution?

 

Strong base problem:  [OH¯] = 2 × [Ba(OH)2] = 2 × 0.055 M = 0.11 M

pOH = –log(0.11) = 0.96

pH = 14.00 – 0.96 = 13.94  (2 digits past decimal)

 

 

 

 

 

14.  (10 points)  Calculate the solubility in grams per liter of CaF2.  The Ksp for CaF2 is equal to 5.3 × 10‑11.

 

Setting up the equilibrium problem:

 

CaF2(s)

ƒ

Ca2+(aq) +

2 F¯(aq)

Initial

–

 

0

0

Change

–

 

+x

+2x

Equilibrium

–

 

x

2x

 

Let x = moles per liter of CaF2 that dissolve.  Then, [Ca2+] increases by x and [F¯] increases by 2x.  Placing the equilibrium concentrations into the Ksp expression and solving for x:

 

Therefore 2.4 × 10-4 mol/L of CaF2 dissolve.  Multiply by the molar mass of CaF2 to obtain grams per liter:

(2.4 ×10-4 mol/L)(78.08 g/mol) =0.019 g/L  (2 sig figs)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

15.  (15 points)  What is the pH of a 0.15 M KCN solution.  The Ka for HCN is equal to 4.0 × 10-10.

 

KCN is a salt.  It dissolves in water to produce K+ and CN¯:

KCN(aq)  ¾¾®  K+(aq)  +  CN¯(aq)

CN¯  is the conjugate base of HCN with a Kb value equal to:

The pH of 0.15 M CN¯ with Kb = 2.5 × 10-5 is a weak base problem:

 

CN¯(aq) +

H2O(l)

ƒ

HCN(aq) +

OH¯(aq)

Initial

0.15

–

 

0

0

Change

–x

–

 

+x

+x

Equilibrium

0.15 – x

–

 

x

x

 

Place the equilibrium concentrations into the Kb expression and solve for x:

 

 

 

pOH = –log(1.9 × 10-3) = 2.71

 

pH = 14.00 – 2.71 =  11.29 (2 digits past decimal)

 

 

(10 points extra credit)  Give the complete definition for 2 out of the 3 terms below.  5 points will be given for each correct complete definition.  Clearly note which 2 definitions you want graded, if you write three definitions without a clear notation, only the first two will be graded.

 

 

Buffer solution: a solution containing roughly equivalent amounts of either a weak acid and its conjugate base or a weak base and its conjugate acid.

 

 

 

 

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

 

 

 

 

 

Amphiprotic:  a substance that can act both as a Bronsted acid and a Bronsted base.