Chem 120A Exam 3 Name:___________________________ IMPORTANT: This exam is designed for you to learn how to apply the concepts you have been practicing. Read the entire question and answer as fully as you can. Be sure to attempt all parts of a question. Always show your work to maximize your points awarded, unsupported answers may not receive full credit and partial answers will receive some partial credit. Circle your answers. All multiple choice have exactly one answer. Write your name on each page in case your exam falls apart during grading. Scratch paper is provided at the end of the exam. Indicate continued work so the graders can give partial credit. Question 1: Solutions multiple choice A – D (3 points each) A. Consider five solutions that all have the same mass of solute in 100.0 mL of solution. Which has the highest concentration as measured in molarity? a) KBr MW KBr = 119 g/mol There are 0.84 moles of K+ and Br‐ 1.68 moles in total + ‐ b) NaF 4.76 moles in total MW NaF = 42 g/mol There are 2.38 moles of Na and F
+
‐ 2.58 moles in total MW KCl = 74.55 g/mol There are 1.34 moles of K and Cl
c) KCl +
‐ and F
3.42 moles in total MW NaCl = 58.44 g/mol There are 1.71 moles of Na
d) NaCl MW CaCl2 = 111 g/mol There are 0.9 moles of Ca+ and 1.8 moles of Cl‐ 2.7 moles in total e) CaCl2 B. Which reaction would not be expected to form a precipitate when dilute solutions of each are mixed? a) AgNO3 + NaCl b) NaCl + KNO3 c) Pb(NO3)2 + CaS d) MgBr2 + NaOH C. Consider pure water separated from an aqueous sugar solution by a semipermeable membrane, which allows water to pass freely but not sugar. After some time has elapsed, the concentration of sugar solution a) will have increased. b) will have decreased. c) will not have changed. d) might have increased or decreased, depending on other factors. e) will be the same on both sides of the membrane. D. A salt solution sits in an open beaker. Assuming constant temperature and pressure, the boiling point of the solution Higher molarity increases the boiling point, a) increases over time. over time the air in the salt solution will b) decreases over time. evaporate, leaving a higher molarity salt c) stays the same over time. d) We need to know which salt is in the solution to answer this. e) We need to know the temperature and pressure to answer this. Stanich – SU12 Page 1 Chem 120A Exam 3 Name:___________________________ Question 2: (8 points) One of the structures below is vitamin C, and the other is vitamin A. Vitamin C is readily excreted in the urine, so it must be constantly consumed to maintain sufficient levels in the body. In contrast, vitamin A is can be stored for much longer periods of time in the body, so it does not need to be replenished as often. Predict which structure is which, and briefly explain your reasoning. H
H
HO
O
C
O
C
C
C
H
C
O
O
C
H
Vitamin C, This molecule has a lot of polarity. It will be water soluble and removed from the body through urine. This makes it necessary to replenish constantly. H
H
H
H
C
OH
H
H
C
H
C
H
H
CH3
C
C
C
C
H
H
C
C
H
H
CH3
H
H
C
C
C
C
C
C
C
H
H
O
H
Vitamin A, This molecule is very hydrophobic. It will be fat soluble and stored in the fat. This makes it unnecessary to replenish as often. H
H
H
H
Question 3: (3 points each) Consider two solutions, A and B, separated by a semipermeable membrane. Fill the blanks below with A (left) or B (right). A. Solution A is pure water, and solution B is 0.05 M glucose. If the membrane is only permeable to water, water will move to the ______B________________compartment. B. Solution A is 0.10 M in fructose and 0.05 M in glucose. Solution B is 0.20 M in sucrose. The direction of osmosis will be to the __________B_____________. C. Solution A is 0.10 M in lactose and 0.050 M in urea. Solution B is 0.15 M in KCl. Water will flow to the ___________B___________compartment. D. Solution A is 0.010 M glucose, and solution B is 0.050 M glucose. The glucose will dialyze to the ____________A_________. E. Solution A is 0.05 M sodium chloride and solution B is 0.05 M calcium nitrate. Water will flow to the _____________B_______________ compartment. Stanich – SU12 Page 2 Chem 120A Exam 3 Name:___________________________ Question 4: (12 points) Ringer's solution, used in the treatment of burns and wounds, is prepared by dissolving 4.30 g of NaCl, 0.150 g of KCl, and 0.165 g CaCl2 in water and diluting to 500. mL. A) What is the percent (m/v) composition of each component? . 100% 0.860% / NaCl: .
KCl: 100% 0.0300% / .
CaCl2: 100% 0.0330% / B) What is the molarity of Cl‐ in Ringer’s solution? . 0.07357
0.07367 mol Cl‐ MW NaCl: 58.44 g/mol NaCl: .
MW CaCl2: 110.98 g/mol MW KCl: 74.55 g/mol .
.
.
.
/
0.001487
/
/
0.002012
0.002974 mol Cl‐ 0.002012 mol Cl‐ Total Cl‐ moles = 0.07866 mol Molarity = 0.07866 mol / 0.500 L = 0.158 M Stanich – SU12 Page 3 Chem 120A Exam 3 Name:___________________________ Question 5: (18 points)Consider the following equilibrium and answer the questions below:
N2(g) + 3H2(g)
2NH3(g)
K = 2.3  10–6
A. The equilibrium concentrations of N2 and H2 are 1.5 M and 2.5 M, respectively. What is the equilibrium concentration of NH3? 2.3
10 2.3
10 2.3
10 1.5 2.5 5.4
7.3
10 10 M B. At equilibrium, the forward reaction rate is: a) Zero b) Greater than the backward rate c) Lower than the backward rate d) Equal to the backward rate C. Fill in the blanks below with “left”, “right”, or “no change”: a) Addition of N2 will cause the equilibrium to shift ____Right_________________________________. b) Reduction of the volume of the reaction vesicle causes the equilibrium to shift ____Right_________. c) Addition of a catalyst to the reaction will cause the equilibrium to shift _________No change______. d) Removing NH3 will cause the equilibrium to shift ____________Right________________________. Stanich – SU12 Page 4 Chem 120A Exam 3 Name:___________________________ Question 6: (3 points each) Consider the energy diagram to the right showing a reaction with and without a catalyst. Fill in the blanks below with the appropriate letter (A, B, C, etc.). a. The activation energy of the catalyzed reaction is represented by ______C______________________. b. The heat of reaction would be represented by ______A______________. c. The lowering of the activation energy by the catalyst is represented by ______B__________. Question 7: Acid/base multiple choice A – F (3 points each) A. Which of the following indicates the most acidic solution? a) pH = 1.5 b) pOH = 5.9 pH = 8.1 c) [OH–] = 0.5 M pH = 13.4 d) [H+] = 0.1 M pH = 1.0 B. The hydrogen sulfate or bisulfate ion HSO4– can act as either an acid or a base in water solution. In which of the following equations does HSO4– act as an acid? a) HSO4– + H2O  H2SO4 + OH–
b) HSO4– + H3O+  SO3 + 2H2O
c) HSO4– + OH–  H2SO4 + O2–
d) HSO4– + H2O  SO42– + H3O+
C. If the following substance is dissolved in pure water, will the solution be acidic, neutral, or basic?
solid ammonium nitrate (NH4NO3) a) Acidic b) Neutral c) Basic D. If the following substance is dissolved in pure water, will the solution be acidic, neutral, or basic?
solid potassium acetate (CH3COOK) a) Acidic b) Neutral c) Basic Stanich – SU12 Page 5 Chem 120A Exam 3 Name:___________________________ E. Which of the following is present in pure water? a) H3O+ b) OH
c) H2O d) All are present in pure water F. Which of the following statements best describes the relationship between conjugate acids and bases? a) as acid strength increases, the conjugate base strength increases b) as base strength increases, the conjugate acid strength decreases c) the conjugate base of a strong acid is always strong d) there is no specific correlation Question 8: (15 points) If 17 mL of 0.54 M HCl is added to 113 mL of 0.28 M NaOH, what is the final pH? Final volume = 17 mL + 113 mL = 130 mL Moles of H+ from HCL: .
Moles of OH‐ from NaOH: 17
.
9.18
113
10 moles 3.164
10 moles There are more OH‐ ions around, so H+ is the limiting reactant: Moles OH‐ left: 3.164 x 10‐2 moles – 9.18 x 10‐3 moles = 0.02246 moles .
.
0.17M is the molarity of OH‐ You can do the next part 2 ways: Way #1: pH = 14‐pOH pOH = ‐log(0.17) = 0.76 pH = 13.2 Way #2: Kw = [H+][OH‐] 1.0 x 10‐14 = [H+] [0.176] [H+] = 5.7 x 10‐14 pH = ‐log(5.7x10‐14) = 13.2 Stanich – SU12 Page 6 Chem 120A Exam 3 Name:___________________________ Question 9: (19 points) You have been asked to make a buffer solution using one of the acids below. A. Fill in the table with the correct conjugate base to each acid. Acid Component HC2H3O2 H2CO3 H2PO4‐ NH4+
Ka 1.8 x 10‐5
4.3 x 10‐7
6.2 x 10‐8
1.8 x 10‐10 Base Component C2H3O2‐
HCO3‐
HPO4‐2
NH3
B. If you want the pH of the buffer solution to be around 6.5, which acid and its conjugate base would you use with equal equilibrium concentrations? (show at least one example calculation) If [A‐] = [HA], then Ka = [H3O+] Then pH = ‐log(Ka) The correct answer is H2CO3 pH = ‐log(4.3x10‐7) = 6.37 C. Using the same acid and conjugate base as in part B, what would the pH of the buffer be if the equilibrium concentration of the conjugate base was 15 times higher than the concentration of the acid? 1
15
2.9
10
pH = 7.54 Stanich – SU12 Page 7 Chem 120A Exam 3 Name:___________________________ Congratulations! You are done. Be sure to check that you have units, correct sig figs and circled answers. Be sure to write your name on each page in case the exam falls apart during grading. You may hand in your exam and leave quietly once you are done UNLESS there is only 5 minutes remaining. Stay seated to be courteous to your finishing classmates. Exams and unreturned materials will be available in BAG 303 starting the first day of the Fall quarter. The key to this exam will be posted on the website tomorrow. Email Cynthia directly if you find a mistake in your Catalyst gradebook, your grade will be calculated directly from those numbers. I really enjoyed teaching you all and wish you luck on your future educational goals! Thispageisforscratchpaper.Besuretolabelcontinuedworkso
thatthegraderswillconsideritforpartialcredit.
Stanich – SU12 Page 8