Department of Chemistry, Geosciences and Environmental Sciences

 

                                                                          Exam 2-A

                       

Chemisty 1084:  Section 010                                                                                  Spring 2008

 

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.  What is the conjugate base of the dihydrogen phosphate ion, H2PO4¯?

            (a)  H3PO4                                                                    (c)  HPO42-

            (b)  H2PO3¯                                                                  (d)  PO43-

 

Answer:  C

To get a conjugate base, remove an H+ from the formula by reducing the number of hydrogen atoms by 1 and reducing the charge by 1.

 

2.  In a neutral aqueous solution:

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

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

 

Answer:  C

Definition of a neutral aqueous solution.

 

3 NO(g)    NO2(g)  +  N2O(g)   DH = –175 kJ

 

3.  Given the reaction above, state how the reaction will respond to the following disturbances by filling in the blank given after the disturbance.  Your choices are:  “move towards products”, “move towards reactants”, or “no effect”.  Each correct answer is worth 1 point.

 

(a)  NO2(g) is removed from the reaction flask                             Move towards products

NO2 is a product.  According to LeChatelier’s principle, removing product will cause the reaction  to replace the removed product by moving in the forward direction.

 

(b)  The total pressure is increased                                               Move towards products

Increasing pressure will favor the side with the lesser number of moles of gas which is the product side.

 

(c)  The temperature is decreased                                                 Move towards products

Decreasing temperature favors the exothermic direction

 

(d)  A catalyst is added to the system                                            No effect

Addition of a catalyst does not effect equilibrium mixtures.

 

4.  In an aqueous solution at 25°C, what is the [H+] when [OH¯] is equal to 6.7 × 10-4 M?

       (a)  1.5 × 10-11 M                                                               (c)  6.7 × 10-18 M

       (b)  6.7 × 10-10 M                                                              (d)  1.0 × 10-14 M

 

Answer:  A

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

       (a)  NH4Cl                                                                         (c)  CsI

       (b)  KF                                                                              (d)  Fe(NO3)3

 

Answer:  B                                                                             

For KF, F¯ is the conjugate base of the weak acid HF so it will produce a basic solution.  Both NH4Cl and Fe(NO3)3 have acidic cations (NH4+ and Fe3+, respectively) so they produce acidic solutions.  CsI has a neutral cation and anion to produce a neutral solution.

 

6.  Which is the strongest base?

       (a)  Ammonia (pKb = 4.74)                                               (c)  Methylamine (pKb = 3.37)

       (b)  Trimethylamine (pKb = 4.20)                                     (d)  Dimethylamine (pKb = 3.32)

 

Answer:  D

For pKb’s, the strongest base has the smallest value of Kb.

 

 

 

7.  The titration curve shown above is representative of which type of titration?

            (a)  Strong acid-Strong base titration

            (b)  Weak acid-Strong base titration

            (c)  Strong acid-Weak base titration

            (d)  Impossible to determine without more information.

 

Answer:  B

This titration curve has a pH at the equivalence point which is greater than 7.0.  It must be a weak acid-strong base titration.

 

8.  Which of the following combinations will form a buffer solution when dissolved in water?

       (a)  HClO4, NaClO4                                                          (c)  NaOH, HNO3

       (b)  HCl, KCl                                                                    (d)  HF, NaF

 

Answer:  D

Only choice (d) has the combination of a weak acid and  its conjugate base.  Choices (a) and (b) are conjugate acid base pairs for strong acids.  Choice (c) is a strong base and a strong acid.

 

9.  Which of the following salts has the lowest solubility in water?

            (a)  PbF2  (Ksp = 7.12 × 10-7)                                        (c)  FeF2  (Ksp = 2.36 × 10-6)

            (b)  BaF2  (Ksp = 1.84 × 10-7)                                       (d)  CdF2 (Ksp = 6.44 × 10-3)

 

Answer:  B

For salts that dissolve to produce the same number of ions, the salt with the lowest solubility is the one with the smallest Ksp value.

 

 

 

 

 


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 0.025 M HNO3?

 

HNO3 is a strong acid solution:  HNO3(aq)  ¾¾®  H+(aq)  +  NO3¯(aq)

[H+] = [HNO3] = 0.025 M

pH = –log(0.025) = 1.60  (2 digits past the decimal place)


12.  (15 points)  The value of c is 0.14 for the reaction:

2 BrCl(g)   Br2(g)  +  Cl2(g)

       Suppose that a reaction flask is filled with BrCl(g) until its concentration is equal to 0.50 M.  The contents in the flask are then allowed to equilibrate.  What are the equilibrium concentrations of Br2, Cl2, and BrCl in the flask.

 

Let x = Δ[Br2],

            then x = Δ[Cl2], and

            –2x = Δ[BrCl]

Add the initial and change in concentrations to calculate equilibrium cocentrations.

 
Initial concentration:  [BrCl] = 0.50 M

Set up an equilibrium table:

 

2BrCl(g)

Br2(g) +

Cl2(g)

Initial

0.50

 

0

0

Change

–2x

 

+x

+x

Equilibrium

0.50 – 2x

 

x

x

 

 

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

Answer the question:

       [Br2] = x = 0.11 M

       [Cl2] = x = 0.11 M

       [BrC;] = 0.50 – 2x = 0.28 M  (2 sig figs in each answer)
13.  (10 points)  Calculate the pH of 0.10 M NH(CH3)2.  The Kb value for NH(CH3)2 is equal to 4.8 ×  10-4.

 

 

Initial concentration:  0.10 M NH(CH3)2.   

 

Since you are given a K­b, this is a weak base problem.  Setting up the weak base problem:

 

NH(CH3)2(aq) +

H2O(l) 

HNH(CH3)2+(aq) +

OH¯(aq)

Initial

0.10

 

0

Let x = Δ[OH¯]

 
0

Change

x

 

+x

+ x

Equilibrium

0.10 – x

 

x

x

 

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

The pOH is equal to:  pOH = –log(6.7×10-3) = 2.17

Converting to pH:  pH = 14.00 – 2.17 = 11.83

 

14.  (10 points)  Calculate the solubility, in grams per liter, of CaF2 in pure water.  The Ksp for CaF2 is equal to 1.46 × 10-10.

 

Let x be equal to the number of moles per liter of CaF2 that dissolve.  Then D[Ca2+] = x and D[F¯] = + 2x

 
Setting up the solubility problem in pure water:

 

CaF2(s)

Ca2+(aq) +

2F¯(aq)

Initial

 

0

0

Change

 

+x

+2x

Equilibrium

 

x

2x

 

Put the equilibrium concentrations into the Ksp expression.

Solve for x:

3.32 × 10-4 mol/L of CaF2 dissolve.  Multiply by the molar mass of CaF2 to get the solubility in grams per liter:

  (3 sig figs)
15.  (15 points)  Calculate the pH of a 0.10 M NaBrO solution.  The Ka value for HBrO is equal to 2.8 × 10-9.

 

NaBrO is a salt which ionizes completely according to the reaction:

NaBrO(s)  ¾¾®  Na+(aq)  +  BrO¯(aq)

BrO¯ is the conjugate base of HBrO with a Kb value equal to:

So you need to calculate the pH of 0.10 M BrO¯ with a Kb value equal to 3.6 × 10-6.  This is a weak base problem:

 

BrO¯(aq) +

H2O(l) 

HBrO(aq) +

OH¯(aq)

Initial

0.10

 

0

Let x = Δ[OH¯]

 
0

Change

x

 

+x

+ x

Equilibrium

0.10 – x

 

x

x

 

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

The pOH is equal to:  pOH = –log(6.0×10-4) = 3.22

Converting to pH:  pH = 14.00 – 3.22 = 10.78


(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 way so that the disturbance is partially counteracted.

 

 

 

 

Amphoteric substance:  a substance that can react with acids and bases.

 

 

 

 

Equivalence point of a titration:  the point at which stoichiometric amounts of acid and base have been mixed in a titration.