exam 3 - 70/70 Quantitative Analysis CHM 201 Exam...

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Unformatted text preview: 70/70 Quantitative Analysis CHM 201 Exam 3 (Chapters 10 and 13) Fall 2012 (70 pts) Name jab W (Wednesday) 1. (10 pts) Calculate the equilibrium constant for the following reaction using the appropriate redox reactions below. 2126‘+ + H3AsO3 + H20 :2: 21:? + Hyaiso4 + 2H+ Fe 2+ + 2c === 126(3) " E°= —0.447 V \flFe3" + 2e" :3. 2Fe2+ E°=O.771V \/H3ASO4 + 2H“ + 2e' 2:: H3A503 + H20 E°=0.560V H3AsO3 + 3H“ + 3e‘ =-.== As(s) + 31120 E°=O.248V 3&3" +/3/. All-"m hr: 0.77lv *- 1+,Asa,+ aw» 9/.” *- H My»; “E‘Uw : mmv 2F33:+ H3 A503 +H30T" “4 QF£S++W HBASOq 4'2”" E. 3 0.9Ilv Cell .015...— alomny 07-‘33‘ ~— K: who-msv .__.__ ”DA—emu? g 2: 33' (We... M 2. (15 pts) A weak acid HA (pKa = 5.00) was titrated with 1.00 M KOH. The acid solution had a volume of 1000 mL and a molarity of 0.100 M. Find the pH at the following volumes of base added; Vb = 0, 5, and 10 mL. Loon " upmor ‘lv 1:34:10.“ 9'5 my base -§' HA £43.. H++A~ WEE—1&3 ~ X9“ "d [Hater "“ “ [HA] fH': “13(Looxlo’a W5“ SGML ‘ MEN; Hue B‘IV‘VQIE‘MG, [1910+ The. Volume Rakdgd i‘, )9 OF \\ PM" M “a Mmb‘m Hm pH =fl¥m _,-—.._______. t I“ “it ”flqujeaue poi/H *Alt HA -——=> 11;” "- '3 Kb ‘_ Kl: : [Hm LOHj _ __>_S__ 7t 253 A-+ H30 "rm—'- HA + 0” (A? w [Bor‘x [4.1” :lt-tngu' {m'hr ___;_ .___..,. W {Marx +>< +>< Wi= W * Win”?- (0-0-5551 @HTI =qflqexio“‘M Kb: £2! ~= [,ooxlo W ‘ 100x104“ , ., km LOOK (0‘9 (NH = "' ‘03 1\q‘y§¥6XJO'6) : E02 ' 7:. ; ,.; H" V4» - 0 = , _, - m jToT {HR 703;... (d{:{;r/‘ P P P H’ I“! 00 3.0? " £3; = (0-loor§(loo.oz “0.0;er [07w 0.010?qu 0 r"— 3. (15 pts) A20.0mLsolution ofadibasicbaseOJOGMBwastitmtcdwifllOJOOMi-ICI. Calculate the pH afier the addition of 5, 20.0, and 35.0 mL ofthe HCI. (Kaa = 1-00 x 10‘ and Kb2= 1.00x 10"“) N" ._._ 6M 1 , m: 10.00 Veilfi) = {DO-omlHO'IDOMW : 90.0.1, VcQMQ = “om" . , 0,+ooM 1“, ._ , , 51.00 ;VCL: EONL 2:339 40 L4 Balkan“ P'Ufl B 1- H" 1&— 9”"- ..___,._.._ Wu 115;,“ “a“! | 7%» _. [mi “22:13— 94.0.9”. 0” : K JV 0 UH’+ . r [8] T7; PH: 9km -Pou = Moo — 3.5a : 842 ,0”: 8-”8 .- "V ”M” ._ mm-‘ a 2 Vm‘: 20.0le K d M e’tu-‘mleuce {‘h“ 3H++V H 062‘ 8 * V 0* 1! MI 9..., 6”? a V. 3 8H4. + ”3° ”if: 6’95 4. OH— POH Q [g (9K5. + (who) a ’5 ( 3-00 + 10 005 f“ r'w ' W = m.oo- 8.00 (you-a 3.00 f": g 00 ‘ fofl- ‘9‘ 9 ;_ :0 l , ‘ and b «up “3:: '° 1 9A1 at p+ 110—)! 40 6W + H+ it; GHa '5' a w I 2.4, __ “Ml 52:50 3—— W%o.o=°-IY {30" = phi + I03 [3H5 TM P0" ~‘-‘ woo + l. o. 3 33} s: l0.‘i§' 9'“ P“ W m _* gm; 9 4 ( 15 pts) The voltage of the ollowing cell“ 18 0.490 V. Find the K, for the organic bfse RN32. -« * ——-——-—s ‘Wx nah-dc K), Pi(s)/H2(1 00 bar)/RNH;(aq 0.10 M), RNH3 Cl (aq, 0 osoM/I SHE RN93) + Hao ..... RN» + d». . -HMF (MNHm5UeJm1mQ + . w_..,.___ E), = [M {HOLD I Cat-«a e. H {m’fl‘Pa-fi AH ”(1(3) Eo : WOOOOV m?“ A“ J91: _H+(q,h2 Mytfifv- A‘tfllfif : o. ooov L“x...,_ \ «in: ”+0 with“ V... in” (7H) Egaooov I 5.. 0'0 Ectu E26” ._._hflifi 105 Q 05430: 0,000 ._ (laws I 7 “*- W3 0.an ,< P?” : =2 [ + 10 “9050196 . Hjin an 3:, [HQ & 5 QkfxlO-fla k: —U++ Lb»? _ (15 pts) Answer the following questions: a. The relationship between the standard potential and formal potential requires what the book refers to as other terms. Without writing the exact equation for other terms indicate what variables are contained in this equation. Describe, briefly (a couple of sentences, be specific), the reason for coming up with both a standard and a formal potential. VMiaueS like. Lira) K‘n and “La. are [mink-rd. in 4;,“ quad-on. The. S‘hmdard. Pol-en \{x'uJ (g9) “0”“ +0 filled-{ms “kg-e W flfl-«e 0‘ “All”; polenlml; were. developed with Hm. 596 19 «it», relevance fun}, i 5': 0.000% The s - QMH“ ‘9‘“ developed 4-0 may,“ values in Pulrwlor In}; ”gel-Em, Jack CM 55 N Cnlwlflu m, Mm}! ”u {Janka} 0F a math/é cells in which Mb! MM we OCCUTIhxa “View *4 E'la -¢ em; we fit «*7. Vote“ PM“: were al I. b‘l ‘I‘DLkWASSH ufcaosgdéflwb 3k: “5?: “:5 9 up Ufunuq vex! do», '1'. 7 g, 9”“: ff Y belies 0 °~ 5* F , - ' 3“ char! g. in threesrifiicegrovmfi raw Ziihcfllatg'e'l?ompletel§° the g‘a \fanic cell‘rgfirem Mllowm I“ Megan}, line notation: ' ‘ m‘xw Zn(s)/Z::Clz(aq)//Cl‘(aq)/Clz(l)/C(s) :7 oi K (100 Q, ‘1: ' . 13 0} . . “MW. A'VVAQ. V C “{hod E/ I c. Balance the following half reactions and give the final overall (balanced) redox reaction: (Hint: The Chromium is in an acidic solution) Cr2012'(aq) + 99,- + Hut 49091“) +1 Mao CF 2” e3+ + e C 09' a» “a 7 6‘1““?- av ”WW-fie 9"361‘" 653+ 7Hao...
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