Department of Chemistry, Geosciences and Environmental Sciences
Exam 2A
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
K_{w} = [H^{+}][OH¯] = 1.00 × 10^{14}
at 25°C
Quadratic Formula: For
an equation of the form: ax^{2}
+ bx + c = 0
_{}
HendersonHasselbach 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, H_{2}PO_{4}¯?
(a) H_{3}PO_{4} (c) HPO_{4}^{2}^{}
(b) H_{2}PO_{3}¯ (d) PO_{4}^{3}^{}
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) _{} NO_{2}(g) + N_{2}O(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) NO_{2}(g) is removed from the reaction flask Move towards products
NO_{2}
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) NH_{4}Cl (c) CsI
(b) KF (d) Fe(NO_{3})_{3}
Answer: B
For
KF, F¯ is the conjugate base of the weak acid HF so it will produce a basic
solution. Both NH_{4}Cl and Fe(NO_{3})_{3} have acidic cations (NH_{4}^{+} and Fe^{3+},
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 (pK_{b} = 4.74) (c) Methylamine (pK_{b} = 3.37)
(b) Trimethylamine (pK_{b} = 4.20) (d) Dimethylamine (pK_{b} = 3.32)
Answer: D
For pK_{b}’s, the strongest base has the smallest value
of K_{b}.
7. The titration curve shown above is representative of which type of titration?
(a) Strong acidStrong base titration
(b) Weak acidStrong base titration
(c) Strong acidWeak 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 acidstrong base titration.
8. Which of the following combinations will form a buffer solution when dissolved in water?
(a) HClO_{4}, NaClO_{4} (c) NaOH, HNO_{3}
(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) PbF_{2} (K_{sp} = 7.12 × 10^{7}) (c) FeF_{2} (K_{sp} = 2.36 × 10^{6})
(b) BaF_{2} (K_{sp} = 1.84 × 10^{7}) (d) CdF_{2} (K_{sp} = 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 K_{sp}_{ }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 HNO_{3}?
HNO_{3} is a strong acid solution: HNO_{3}(aq) ¾¾® H^{+}(aq)
+ NO_{3}¯(aq)
[H^{+}] = [HNO_{3}] = 0.025 M
pH = –log(0.025) = 1.60 (2 digits past the decimal place)
12. (15 points) The value of K_{c} is 0.14 for the reaction:
2 BrCl(g) _{} Br_{2}(g) + Cl_{2}(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 Br_{2}, Cl_{2}, and BrCl in the flask.
Let x = Δ[Br_{2}], then x
= Δ[Cl_{2}], 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) 
Br_{2}(g) + 
Cl_{2(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:
[Br_{2}] = x = 0.11 M
[Cl_{2}] = 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(CH_{3})_{2}. The K_{b}
value for NH(CH_{3})_{2} is equal to
4.8 × 10^{4}.
Initial concentration: 0.10 M NH(CH_{3})_{2}.
Since you are given a K_{b}, this is a weak base problem. Setting up the weak base problem:

NH(CH_{3})_{2}(aq) + 
H_{2}O(l) 
HNH(CH_{3})_{2}^{+}(aq) + 
OH¯(aq) 

Initial 
0.10 
– 

0 
Let x = Δ[OH¯] 

Change 
– x 
– 

+x 
+ x 

Equilibrium 
0.10 – x 
– 

x 
x 
Place the equilibrium concentrations into the K_{b} 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 CaF_{2} in pure water. The K_{sp}_{ }for CaF_{2} is equal to 1.46 × 10^{10}.
Let x be equal to
the number of moles per liter of CaF_{2} that dissolve. Then D[Ca^{2+}]
= x and D[F¯] = + 2x
Setting up the
solubility problem in pure water:

CaF_{2}(s) 
Ca^{2+}(aq) + 
2F¯(aq) 

Initial 
– 

0 
0 
Change 
– 

+x 
+2x 
Equilibrium 
– 

x 
2x 
Put the equilibrium concentrations into the K_{sp} expression.
Solve for x:
_{}
3.32 × 10^{4} mol/L of CaF_{2} dissolve. Multiply by the molar mass of CaF_{2} 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 K_{a} 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 K_{b} value equal to:
_{}
So you need to calculate the pH of 0.10 M BrO¯ with a K_{b} value equal to 3.6 × 10^{6}.
This is a weak base problem:

BrO¯(aq) + 
H_{2}O(l)


HBrO(aq) + 
OH¯(aq) 

Initial 
0.10 
– 

0 
Let x = Δ[OH¯] 

Change 
– x 
– 

+x 
+
x 

Equilibrium 
0.10
– x 
– 

x 
x 
Place the equilibrium concentrations into the K_{b} 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.