Name: Grading key Page 1/9. For use by grader Page Points Score

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1 Name: Grading key Page 1/9 Chemistry 125/126, Exam 2 Tuesday, April 20, 2010 Welcome to the first hourly exam for Chemistry 125/126. This exam consists of 7 questions worth a total of 75 points plus a bonus question worth 3 points for a possible total of 78 points. It is 10 pages long; 8 pages of questions including a bonus question and periodic tables (page 10). To receive credit, your answers must be placed in the indicated spaces or boxes. If calculations are required to obtain an answer, show your calculations; you will not receive credit for numerical answers alone. Do not write in any box marked "For use by grader". The exam should take about l hour, however, you may use up to 1.5 hours to complete the exam. All exams must be turned in at 7:45 pm. Save any questions you have during the exam for Prof. Kerner. The GSIs have been told not to answer questions during the exam. Graded exams will be available starting tomorrow. If, after checking the exam key, you feel there was a grading mistake, turn your exam in with a statement describing the mistake to Prof. Kerner s mailbox in 1500o chem. All re-grades must be turned in by noon, Monday, April 26, Course Information Section GSI Section GSI Section GSI 127 T AM Kevin Hartman 191 T PM2 Zhenxin Lin 237 Th PM2 Blake Erickson 129 T AM Rahul Rattan 195 T PM2 Shuwen Sun 239 Th PM2 Russell Bornschein 131 T AM Lauren Soblosky 199 T PM2 Heidi Pedini 241 Th PM2 Joshua Skodack 135 T PM1 Zhenxin Lin 233 ThPM1 Lauren Soblosky 251 F PM1 Akiko Kochi 137 T PM1 Rahul Rattan 225 Th PM1 Shuwen Sun 253 F PM1 Kevin Hartman 139 T PM1 Rachel Barnard 227 Th PM1 Blake Erickson 119 F PM2 Akiko Kochi T PM1 T PM2 Russell Bornschein Joshua Skodack Th PM1 Th PM2 Rachel Barnard Heidi Pedini 123 F PM2 Alaina DeToma For use by grader Page Points Score Name: GSI: Section: (+03) TOTAL 75 (+03)

2 Name: Grading key Page 2/9 Question 1 (15 points) requires you to use your knowledge of structure and properties and periodicity to determine the identity of chemicals. Periodic tables are provided at the back of the exam for your reference. A. A white compound dissolves in water (ph 7) and the resulting solution tests acidic. Circle the white compound testing acidic in water: Na 2 CO 3 Al(NO3)3 Zn(OH) 2 B. You need to distinguish between solid white compounds. Tests that you can perform are: 1. Determine if the compounds dissolve in 1M NH Determine if the compounds are soluble in water. 3. Determine the ph values of aqueous solutions of the compounds. A desirable test produces notably different results for the two compounds. Select the number (1, 2, or 3) for the one best test for distinguishing between the two compounds. White Compounds Test (1, 2, or 3) BaCO 3 vs ZnCO 3 1 Ca(NO 3 ) 2 vs Pb(NO 3 ) 2 3 AgNO 3 vs AgCl 2 each C. Predict the comparative base strengths of RbOH, Sr(OH)2, and Ga(OH)3. Strongest base Weakest base RbOH > Sr(OH)2 > Ga(OH)3 D. Predict the comparative acid strengths of HBrO 3, HIO 3, and HClO 3. Strongest acid Weakest acid HClO3 > HBrO3 > HIO3

3 Name: Grading key Page 3/9 Question 2 (16 points) deals with a study of acids and bases and their measured ph values: Sample ph Sample ph 0.10 M pentanoic acid M nitric acid (HNO 3 ) 1.0 (C 4 H 9 COOH or HC 5 H 9 O 2 ) 0.10 M sulfurous acid (H 2 SO 3 ) M ammonia (NH 3 ) M boric acid (H 3 BO 3 ) M potassium hydroxide (KOH) 13.0 A. Based on the ph data compare the acid strengths of pentanoic, sulfurous, and boric acids. strongest acid sufurous > pentaonoic > boric weakest acid B. The electronegativity of phosphorous (P) is greater than the electronegativity of boron (B). Predict the comparative ph of 0.10 M H 3 PO 3 and 0.10 M H 3 BO 3. Indicate (circle) if the ph of 0.10 M H 3 PO 3 is Less than equal to greater than ph 4.10 (ph of 0.10 M H 3 BO 3 ) C. Based on the ph data choose the reaction below that best describes the reaction of pentanoic acid with water. Place an X by the best reaction. (2 points). For the reaction that you choose, label each species as an acid or a base, according to how it reacts and draw lines connecting each conjugate acid-base pair. (2 points) Best Reaction HC 5 H 9 O 2 + H 2 O OH - + H 2 C 5 H 9 O 2 + HC 5 H 9 O 2 + H 2 O OH - + H 2 C 5 H 9 O 2 + X (2 pts) HC 5 H 9 O 2 + H 2 O H 3 O + + C 5 H 9 O - 2 Acid BASE ACID Base(2 pts) HC 5 H 9 O 2 + H 2 O H 3 O + + C 5 H 9 O - 2

4 Name: Grading key Page 4/9 D. You titrate a sample of pentanoic acid (C 4 H 9 COOH or HC 5 H 9 O 2 ) with 0.10 M NaOH. What is the formula for the salt product? Is the salt product acidic, basic, or neutral? Formula for the salt product: C4H9COONa or NaC5H9O2 Indicate (circle) if the salt product is: acidic basic neutral E. Based on the above ph data, will 0.10 M KHSO 3 or 0.10 M NH 4 HSO 3 have a lower ph? Indicate (circle) the 0.10 M salt solution with the lower ph: KHSO 3 NH4HSO3 Question 3 (3 points) requires you to use your knowledge of Lewis acids and bases to explain the acidity metal ion iron (II). You add Fe(Cl) 2(s) to water to form a clear 0.10 M solution. The picture below represents the Lewis acid iron(ii) ion reacting with a sample of the Lewis base water: Fe O H H O H H? products Circle the correctly completed and balanced equation showing the products in the clear,, acidic solution. 1. [Fe(H 2 O)] 2+ + H 2 O [Fe(OH)] 2+ + H 3 O + 2. [Fe(H 2 O)] 2+ + H 2 O [Fe(H 3 O)] 2+ + OH - 3. [Fe(H2O)]2+ + H2O [Fe(OH)]+ + H3O+ 4. [Fe(H 2 O)] 2+ + H 2 O [FeO] + 2H + + H 2 O 5. [Fe(H 2 O)] 2+ + H 2 O [Fe(OH)] + + H + + OH -

5 Name: Grading key Page 5/9 Question 4 (7 points) deals with your titration of g of an amino acid with the molecular formula C5H9O4N. The end point occurs when you have added ml of M NaOH. You determine that the equivalent weight of the amino acid is A. How many ionizable protons are there per molecule of the amino acid C5H9O4N? For credit be sure to show your calculations. Ionizable protons: = Molecular Wt. = 5 (12) + 9 (1) + 4 (16) + 14 = 147 (1 pt for correct MW; Equivalent Wt pt for dividing by EW) Ionizable protons = 2 (1 pt. for correct calculation) # Ionizable protons/molecule = = 2 3 points B. Your titration results are correct. Your teammate conducts an identical titration (using 0.202g of the amino acid and M NaOH) but her results are NOT correct. Error: Your titration was conducted using the indicator phenolphthalein (ph end point = ph ). Your teammate used methyl red (ph end point = ph ). 1. Determine the impact of the indicator error choice on the volume of NaOH used by your teammate to reach the end point. Indicate (circle) the volume of NaOH used by your teammate at the end point. more than equal to less than ml 2. Determine the impact of the indicator error choice on your teammates calculated equivalent weight value. Indicate (circle) the effect of the error on the calculated equivalent weight of your teammate. greater than equal to less than (eq.wt) 2 points ONLY awarded for greater than choice if less than circled in box above. Any other combos constitute a guess and lack of knowledge/understanding so no credit

6 Name: Grading key Page 6/9 Question 5 (13 points) deals with the reaction of 0.10 M CoCl 2 with 0.10 M Na 2 CO 3 and 1.0 M NH 3 where the following observations are recorded: Step 1. CoCl 2(aq) + Na 2 CO 3 (aq) red precipitate (ppt.) Clear red Step 2. Step 1 mixture + NH 3(aq) red ppt. dissolves; clear yellow solution forms Information: Co 2+ (aq) = [Co(H 2 O) 6 ] 2+. A. When 0.10 M Na 2 CO 3 is added to 0.10 M CoCl 2 a Lewis acid-base reaction occurs resulting in the formation of a red precipitate. The Lewis base species reacting to form the red precipitate = CO3 2- The Lewis formula for the red precipitate formed when Na 2 CO 3(aq) reacts with CoCl 2(aq) = [Co(H2O)5CO3] B. In step 2 (above), the addition of NH 3 (aq) causes the red precipitate formed in step 1 to dissolve. Record the formula for the species (Lewis acid or base) in the step 1 equilibrium system reacting with the NH 3 (aq). The species reacting upon addition of NH 3 (aq) = [Co(H2O)6]2+ C. What will you observe if you add drops of 1 M HNO 3 to the step 2 product mixture (yellow solution) and stir till an excess amount of 1 M HNO 3 has been added? Indicate the correct sequence of observations you record (circle a single choice next to each of steps 3, 4, and 5) as you add drops of 1 M HNO 3 to the step 2 product mixture till excess HNO 3 is added. Step 3. The solution remains yellow A red ppt. appears The yellow solution turns clear red Step 4. The solution remains yellow The red ppt dissolves The solution remains clear red Step 5. The solution remains yellow The solution turns clear red The solution remains clear red

7 Name: Grading key Page 7/9 D. Indicate in five words or less what you will visually observe if you reverse the order of addition of reagents to the of 0.10 M NiCl 2? Note that only the first five words of each recorded observation will be read and graded. Visual observation? Step 6. CoCl 2(aq) + NH 3(aq) Red ppt dissolves to yellow soln# Clear red (1 point or 0) or ( red solution turns yellow )# Step 7. Step 6 mixture + Na 2 CO 3(aq) The solution remains yellow( 3 points)* ( or no change if step 6 correct = 3 pts) ( or no reaction if step 6 correct = 1 pts) If student indicates no reaction it is true but that is not an observation and thus = only 1 pt # observations depend on if cobalt chloride is added to ammonia or ammonia is added to cobalt chloride *Only the first five words of each recorded observation are to be read and graded 4 points Question 6 (6 points) deals with your analysis of a reaction where you add the metal Pd to a yellowgreen solution of 0.10 M FeCl 2(aq) and observe no apparent reaction. Circle every observation (1and/or 2 and/or 3) that can help confirm that NO reaction took place. Observation 1. Add hexane to the reaction mixture and the hexane remains clear and. 2. Add the metal Fe to 0.10 M PdCl 2 and observe a reaction. 3. Add 0.10 M FeCl 3(aq) to 0.10 M PdCl 2 and observe no reaction. For use of grader only 6 points for circling 1 and 2 and NOT circling 3 Each correct observations (1, 2, and 3) = 2 points for each

8 Name: Grading key Page 8/9 Question 7 (15 points) deals with the analysis of the reaction between CuSO 4(aq) and SnCl 2(aq) : CuSO 4 (aq) + SnCl 2 (aq) loss of blue color and formation of a white precipitate. blue Reference Tests 1. CuSO 4 (aq) + NaCl (aq) no reaction. blue 2. CuSO 4 (aq) + Sn(NO 3 ) 2 (aq) loss of blue color. A clear, solution forms. blue 3. Na 2 SO 4 (aq) + SnCl 2 (aq) no reaction. A. What does test 2, by itself, tell you about the reaction of CuSO 4 (aq) and SnCl 2 (aq)? What do you know about the CuSO 4 (aq) and SnCl 2 (aq) reaction from test 2? Cl- is critical to the formation of the white precipitate No credit if the ion Cl - not recorded regardless of the interpretation; 1 pt if ion Cl - recorded but answer does not include fact that it is critical to formation of the white precipitate e.g. if answer only indicates it is a critical species. 3 points B. What does test 3, by itself, tell you about the reaction of CuSO 4 (aq) and SnCl 2 (aq)? What do you know about the CuSO 4 (aq) and SnCl 2 (aq) reaction from test 3? Cu2+ is critical to the reaction (or loss of blue color and formation of white ppt) No credit if the ion Cu 2+ not recorded C. What is the identity of the white precipitate formed when CuSO 4 (aq) and SnCl 2 (aq) react? 3 points The formula of the white precipitate product = CuCl or (CuCl or SnCl4) (3 points) SnCl 4 only answer = 1 point D. What is the identity of the reacting species causing the loss of blue color? Note: For credit do not record any spectator species! The reacting species are = Cu2+ and Sn2+ (both species must be recorded and correct)

9 Name: Grading key Page 9/9 Based on the reaction, what are the relative strengths of Cu 2+ vs. Sn 4+ as oxidizing agents? Strongest Weakest Oxidizing Agent Cu2+ > Sn4+ BONUS QUESTION (3 points) Baking powder contains sodium bicarbonate, NaHCO 3, and the double sulfate salt, NaAl(SO 4 ) 2. When added to moist dough, small bubbles of CO 2 give the finished baked product a light and porous texture:? + H 3 O + CO 2(g) + 2 H 2 O 1. Fill in the missing species in the equation for the reaction of baking powder: HCO3 - + H 3 O + CO 2 (g) + 2 H 2 O 2. Identify the species in the baking powder that is critical to the production of the hydronium ions, when the baking powder is added to moist dough: Al3+(aq) or [Al(H2O)6]3+ (both answers must be correct for credit)