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Exam_I_Key - CHEMISTRY 41 EXAM I SPRING 2006 Name K 8 5 A 4...

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Unformatted text preview: CHEMISTRY 41 EXAM I SPRING 2006 Name: K 8 5/ . A 4 I Date: Show all your work and clearly state any assumptions you make in a problem! No credit will be given for simply providing an answer. It is also essential that you clearly show your set up. For non-calculation based questions, make sure you clearly and completely answer the question. Go through the exam and work those problemsyou know how to do first, then start on the more difficult problems. An equation sheet has been provided at the end of this exam. Only those equations that I told you I would provide you with are on this sheet. You are not permitted to use scratch paper. The back of each page has been left blank for this purpose. If you need more space to answer a question, use the back of the page. If you have forgotten your calculator you will have to do your math long hand. You are not allowed to borrow your neighbor’s calculator. PLEDGE: I have not given nor received aid on this exam, nor have I seen anyone else give or receive aid on this exam. Signature: Name: \Z‘ ej Chemistry 41 1. (28 points) Answer the Following Questions 3. (5 points) Calculate the isoelectric point for a 0.100 M solution of Lysine, the structure of which is shown below. 0 _. B (:zl—OH pKa (a) = 2.04 fif~_3 +H3N—C—H PK?» ([3) =9-08 f K \4 . ‘ CH2 pKa (y) = 10.69 "’ ._ EWQD‘H) ”:3. ('11 ’3‘ f 1 1:1. 9.o%+1o.éo9 THE P 2 : Clgg (l3H2 W: Y Lysine NH3+ b. (4 points) Define amphiprotic §?e C160 A \q Qj‘ C 90 Q“ ”j c (M (“100 Q'lc Q Payton P c. (3 points) Write a charge balance equation for a 0.0500 M NaH2P04 solution. CNN} + [11*] : 111121304] + 21311121111 + 301.113] d. (3 points) Write a mass balance equation for a 0.500 M H2804 solution. 0. 500M : Case/{y [51.3qu 143504 13 s. SMDS QC\d. e. (5 points) 0.100 mL of a 2.50 x 10’5 M HCl solution is diluted to 1000 mL in water. What is the pH of the resulting solution? 6 + _ (Y\ —q M Lin: @3323, Wm“) We 9 5c 0 M >l£ii€l \(33CDrnL. venj CH Vile f. (5 points) A solution contains 75 different conjugate acid- base pairs. Among them 1s acrylic acid and acrylate ion, with the equilibrium ratio [acrylate 1on]/ [acrylic acid]: 0.75 What IS the pH of the solution? C QC rx [li‘ic‘i H2C=CHC02H PH’ PKG “L L03” [cm/1 L QC 1] Acrylic Acid (pKa=4.25) 13H 4 Q 5 + L O: O 75 ' g. (3 points) If the pH of a phosphoric acid (H3PO4) solution 1s adjusted to 8.00, what is the principle species? K31: 7.11 x 10'3 ,=K32 6. 32 x 10 3 ,and Ka3 = 7.1 x 10 13. Pbfl ’2 \5 quZt'llo 9K9? 1.15 Pkcu< P H < ‘PKC‘B 50 Umi,3_\] 2. (15 points) How many mL of 0.500 M NaOH should be added to 0.064 moles of a monoprotic acid (Ka = 8.32 x 109) to give a solution with a pH of 7.60 in a final volume szso'omL? HR + OH“ 2? A“ + H19 \0‘31(?Q€\‘ d.) T QOb‘imM o O / \0'3cl /“\ - E, (9 09444 +>< A, O 28— PH: Pkg + L03 WA 9 7.020: O'OCPDJCLOS 0,0(04’X .2: 03?) : cocoa—x 2) 0.0%} ~o.’33x: X 0.0% = NH X X : mam/3 OH': 01mg; mad) 0.0%ngK me fl . 0.5osman" O'OgQL @3300 3. (10 points) Consider the titration of 100 mL of a 0.100 M solution of NH3 (a weak base) with 1.00 M HCl. a. Sketch what the titration curve would look like and correctly label the axis. 4' l b. Indicate the major regions in this titration and what specie(s) will dictate the pH at each major region. Just provide chemical species, no explanations or rxns. c. Indicate at what point on the below curve the maximum buffer capacity is 4,8 (4‘) reached. Be specific. + \ Cover ed \0 Cl Q35 “#1 NH3 ) NH,‘ 91% 3) Covered. ~(\ (‘52de ~% \2-4 m \2—11 in NH} / (in: an 3 N44“ “Pie; pil- {“4 mm HQ (+29%) W 5mL- \O m\_ Vim} VHu (mq Met/x (Sofia CWQ‘W 4. (20 points) Calculate the pH and equilibrium concentrations of ALL the different species in a 0.010 M potassium carbonate (K2C03) solution. K31 = 4.45 x 10'7 and Kaz = -11 , 4.69x10 . L03:— _’\’\/\&0 F3 HCOE + OH,” Kb; M ; LENS HCOE H410 25 HQCOg + 0H” [Olfl ’7‘ \YfiKb‘ Vb :. \5(DBMS40‘)((_>.<_)lo‘) : M4 NOJM. '\.4>§\<_>3M > (\»\/\\:51\)(Q.-<>\QM\) Roprmt mi dqld. \( ________ L LOKLQ‘A‘ -4 Cow] #(0 4 2. 2-\3><\<> : Cfifimj 9 2.0m) #2-\3>»<\och-3;mw] ’L '4 . [QM *1-‘3 “‘5 fifty/if} ~11? xm : o EMMY. n A (9qu c3 auburn 11-4 @qu d _ , é ’ 3r \\b\+ \\&Q A "\’\’\ (‘4qu Und\55dc‘c‘:\C’C.\ 5. (15 points) A 0.03 M of coulombic acid, a weak acid (HA), is 6 % dissociated. Find the K3 for this acid. [/31 (o oQMoogt/o— \.%><\63w\~[t\+1 Um") (OW/13003111): 0 091.1 M , mkal’vhmooh 4. - -3\2 ,_-_.-.,_,._...1~ m class KC: Cm 7 11.03.2133 -\\_®. >4\© 4\?nsblem 1043 [HA] 0.01%2 ’ ~ 6. (12 points) Answer the Following Titration Questions: 3. (6 points) 30.0 mL of a O 200 M solution of NaHsz4 IS being titrated with 0.300 M KOH. Calculate the pH after the addition of 20 mL KOH. For H3PO4: Ka1= 7.11 x 10 ',3 Ka2 =6. 32 x 10 3 ,and Ka3 = 7.1 x 10_ ".33 (‘30 o mL\(ofLOQ M) - Wm (0.3.31)th \ Pd S‘l exam) Vl :>l_l_90041\}€(b\7.10mL \lngzz‘lOmL 2 20C). «Amok LQ¥ u: r\ c 3* 3“2 4 3“ lebxms x1432 11-1415 [H Po 43:2] (0 :%<3\.\(O1QQH)_ ‘ W»-.. o.\ 1 M 30L + 0 OIOL. [H W K3K3KO ‘3 K2 Kw 8303313331“ Wo‘33l (_o \1\_+<c31x\o 3011'”) ””33: :35”— 3 \‘ (agl3flO c3) + OJ”). EHQ: (1.14 X\a_3"OM ‘ K + K, \ @333 PH” P 12*) 3:91.? \‘1______i/75de§ CbszogK b. (6 points) A 45.0 mL solution of 0.150 M butanoic acid (Ka = 1. 52 x 10'5 CH3(CH2)3C02H) lS titrated with 0.100 M KOH. Find the pH at the equivalence point. K4 ‘1 0 mm (o \ «3M3: \lecb(o loo M3 \IQOB‘ (07- 5 ml. K Kw C (ox ,IO ‘— *’ ‘ l0 ’33 wauw \31 thQ base, A . ‘3 KG _ (0 75X\Oj ml EM: ——»-—3~——-*1"-”“ = ONOLQOM [OH] (3. 04: 3.3 00615 L . #-——~—~—- - *(0 [OI-Ht \lflpbmo 3( 1.1090143) = can.) M 40 H . (6.3km M {Guillooborfi ?fbb\Lm5‘. \7-‘6 V ‘9 and worked. / \: “l. I“, ‘\_ . 'L F“ POK 0:) .313) 1&0 in Class PH: V3. 00 L10 : 88ml ““ ...
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