Exam 4 review - CEM 383 — Fall 2007 Exam 4 Review —...

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Unformatted text preview: CEM 383 — Fall 2007 Exam 4 Review — Chapters 10 — 11 Useful Equations Chapter 11 — Acids and Bases Kw= [H+][OH'] = 1 x 10'14 at 298 K Kw = KaKb pH = 'long+l pOH = -log[H+] pH + pOH = pKW = 14.00 Buffers: pH = pKa + log [base] Range: pH = pKa -l_- 1 [acid] Chapter 10 — Electrochemistry o o o o o ETOT = — E cell = E cathode ' E anode E oxidation = "E reduction nTOT ArGo = 'V FEocell Eoce” = K = VF RT 0 RT ’1‘ 0 0.026 Ecell = Ecell ‘7'?ng at 298 K, Eoell = cell — "‘0 Review Problems 1. Write the dissociation reactions for phosphoric acid (H3PO4) and give the pKa for each. Ka1=7.11x10'3 Ka2=6.32x10'8 Kag,=4.5x10‘13 WW mm" «9912' MW i’lriiwém -: Ht 4. till/"3513"" rim, 3 2.5 HZ fioq' "63 H, “" J H i730“ V pKa; : 7.10 Twat???” a. Ht *4 W»; 5"" PKM W135 t ‘I.< At pHfitO', what is the predominant species in solution? H, P04 1 " Calculate the pH of a solution of 0.210 M K2HPO4 and 0.260 K3PO4. / “Do a) “t 4 HPO 1' Hi“ ‘4 K 1, "’ ‘ ~> mam l 10%; 7) 3i?” WM” 'J‘WW‘“ t 19?! {a “t if?” 1‘ T “t 5‘ “'“W‘ PH l 2 J t ll”2m l J 0le 2. Consider the titration of 35.00 mL of a 0.080 M acetic acid solution with 0.100 M KOH. («A 3 L19 X '02 a. What is the initial pH of the solution? Cfisrozh '3‘ “‘2‘502." “M441; Kat: (CH3C02’7CHU x2 or 8+ o. 0% a o _ : _ 119on 6"] 6-" 0%“- X x x [0”; 601%] o. 0:64 x : 0.00llZZ)o.QOll‘l‘—ll goon-M :CH’r] pH : 2.43 b. What volume of KOH is required to reach the equivalence point? What is the total volume of the solution at the equivalence point? at eq pr rim W5 4’1le 1: mqu basic L t l .. r V :2‘8VHmol m 3mm, 2mm] mm W. M: V10». 3 35" 2? “1' US mill c. What is the analytical concentration of acetate ions at the equivalence point? ' Cuzco2 H + KOH —-) CH3602' + 54+ + Hz 0 all CHggoll—i com/«(Mfd +1» (“3692" c - [gm/Hanoi fie-ms” aha).— - —-——————W ML :wfiwtlm d. What is the pH atthe equivalence point? ,. mi. K 043072." 4 H20 75: (Hyfljfli ‘l OH' Kb : » XV :g‘lpie 30:: y fly )C: (CH? CG; j ' Km ’\ t «O y,” I I ‘ X > PC“; l0"'0 : : r (04/ :CoH- 5‘” Y W” 3 e. What is the pH at the half—equivalence point? ‘I'; o-C CHZCOZE Corn/{vwd «Lao CH3C01’ 9’0 [Cchazii] : Regal—j Echvfl [Eezgc alt-ill f. What is the pH when 34.00 mL of the H solution have been added? Nit f’i’fgtli‘vau’ “(11: pmflf “KOH : 01000 mmol VLle - Wm“): ' l 5 ML — Ccholl-l a KOH MP (HYJCGQ 4- la + H10 2% 3.4 0 o 0 E 010 2.9 2% 2.2” KOH «a K-l fa, CH ' _ ()len/1M k : ’ Memo: 90“ : “SEW: » 2.0V 2?” liflLll 3. What voltage would you expect to measure for the ?ollowing lectrochemical cell? 0. 0€0 WWW ol PH : pm “1’93 M9(3) l M9304 (“W H CUSO4 (1M) | CU(S) Cu2+(aq) + 2e' <—> Cum E = +0.342 v ¥ MW? l‘ .- Mgz’yaq) + 2e‘ H Mg(s) E° = -2.372 v WW WW Eceu T“ EC ’ E“ natured "\ ‘O-JLllv1-2.3410)», _.,. ‘ECCH : 2W4 VI ...
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Exam 4 review - CEM 383 — Fall 2007 Exam 4 Review —...

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