63 K a 31 x 10 8 HX H 2 X 2 Initial 1002 050 Change 2X X X Equilibrium 050 2X X

63 k a 31 x 10 8 hx h 2 x 2 initial 1002 050 change

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(Ka= 3.1 x 10-8)HXH2X2Initial1.00/2 = 0.5000Change-2X+X+XEquilibrium0.50-2X+X+XX2/(0.50 - 2X)2= 3.1 x 10-8X/(0.50-2X) = 1.76 x 10-4X = 0.88 x 10-4= [H2] = [X2]0.50 = [HX]2002 QuizCHEM162-2003 5THWEEK RECITATIONCHAPTER 14 - ACID AND BASE EQUILIBRIAACID AND BASE EQUILIBRIA CALCULATIONS3. Consider the reaction: 2HI(g) H2(g) + I2(g)A container originally holds 0.50 M HI. Suppose 10.0% of this HI reacts as in the above equation, to reach equilibrium. Calculate the value of K for this reaction.HIH2I2Initial0.5000ChangeEquilibriumHIH2I264
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Initial0.5000Change-0.05+0.025+0.025Equilibrium0.450.0250.025([0.025] x [0.025])/[0.45]2= 3.1 x 10-3= KCHEM162-2003 5THWEEK RECITATIONCHAPTER 14 - ACID AND BASE EQUILIBRIAACID AND BASE EQUILIBRIA CALCULATIONS15.23Calculate the pH of each of the following solutions.a. 0.100 M propanoic acid (HC3H5O2, Ka= 1.3 x 10-5)(a) HP + H2O H3O++ P-HPH3O+P-Initial0.10000Change-X+X+XEquilibrium0.100 - X+X+XX2/(0.1-X) = 1.3 x 10-5X = 1.140 x 10-3pH = -log (1.140 x 10-3) = 2.94Chapter 1465
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4.CHEM162-2000 HOURLY EXAM IICHAPTER 14 - ACID AND BASE EQUILIBRIAACID AND BASE EQUILIBRIA CALCULATIONS2 H2O(l) H3O+(aq) + OH-(aq)Kw= 1.0 x 10-14at 25CKw= 3.8 x 10-14 at 40CWhich statement is true concerning pure water at 40C?A.pH = 7.00pOH = 7.00B.pH = 6.71pOH = 6.71C.pH = 6.71pOH = 7.29D.pH = 7.29pOH = 6.71E.pH = 7.00pOH = 6.71Kw= [H+][OH-] = 3.8 x 10-14X = [H+] = [OH-]X2= 3.8 x 10-14X = 1.95 x 10-7= [H+] = [OH-] pH = -log[H+] = -log(1.95 x 10-7) = 6.71pOH = -log[OH+] = -log(1.95 x 10-7) = 6.718.CHEM162-2000 HOURLY EXAM IICHAPTER 14 - ACID AND BASE EQUILIBRIAACID AND BASE EQUILIBRIA CALCULATIONSWhat is the pH of a solution prepared by dissolving 0.80 grams of NaOH (molar mass = 40.0 g/mol) in enough water to make 3.0 liters of solution?(0.80 g/40 g/mol-1)/3.0 L = 0.00667 M NaOH = 0.0067 M OH-pOH = -log [OH-] = -log [0.00667] = 2.18pH = 14.0 - 2.18 = 11.82A.10.89B.11.37C.11.82D.12.30E.10.5266
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9.CHEM162-2000 HOURLY EXAM IICHAPTER 14 - ACID AND BASE EQUILIBRIAACID AND BASE EQUILIBRIA CALCULATIONSWhat is the pH of a 0.50 M aqueous HNO2solution?Ka= 4.3 x 10-4 for HNO2A.3.37B.3.67C.4.67D. 1.84E.2.3710.CHEM162-2000 HOURLY EXAM IICHAPTER 14 - ACID AND BASE EQUILIBRIAACID AND BASE EQUILIBRIA CALCULATIONSNaY is the sodium salt of hypothetical weak acid HY. If a 0.10 M aqueous NaY solution has pH = 11.00, then what is the value of Kafor HY?Y-+ H2O HY + OH-pH 11 = pOH 3 [OH-] = 1 x 10-3Y-H2OHYOH-Initial0.1000Change-1 x 10-3+1 x 10-3+1 x 10-3Equilibrium0.1 - 0.001 = 0.11 x 10-31 x 10-3([HY][OH-])/[Y-] = Kb’([1 x 10-3][1 x 10-3])/[0.1] = Kb’1 x 10-5= Kb’Kw= Kax KbKa= Kw/Kb’Ka= (1 x 10-14)/(1 x 10-5)Ka= 1 x 10-9A.1.0 x 10-5B.1.0 x 10-3C.1.0 x 10-7D.1.0 x 10-9E.1.0 x 10-1167
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12.CHEM162-2000 HOURLY EXAM IICHAPTER 14 - ACID AND BASE EQUILIBRIAACID AND BASE EQUILIBRIA CALCULATIONSHydrazine, N2H4, is a weak base with Kb= 1.7 x 10-6. What is the percent protonation of N2H4in a 0.010 M aqueous solution of hydrazine?A.2.1 %B. 1.3 %C.0.54 %D.2.8 %E.0.85 %N2H4+ H2O N2H5++ OH-N2H4+H2O N2H5+ +OH-Initial0.01000Change-X+X+XEquilibrium0.010 - X+X+X([X] x [X])/(0.010 - X) = 1.7 x 10-6((([X] x [X])/(0.010 - X) = (1.7 x 10-6)),X)X = 1.30 x 10-4mole percent protonation = ((1.30 x 10-4)/0.010) x 100 = 1.368
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14.CHEM162-2000 HOURLY EXAM IICHAPTER 14 - ACID AND BASE EQUILIBRIAACID AND BASE EQUILIBRIA CALCULATIONSIt is found that 25.0 mL of 0.200 M aqueous HCl solution is completely neutralized by 20.0 mL of aqueous Ba(OH)2solution. What is the molarity of the Ba(OH)2solution?
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