FinalExam sample problems_key_081711

FinalExam sample problems_key_081711 - CONCEPTS. 3 pts each...

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Unformatted text preview: CONCEPTS. 3 pts each 1. Which of the following compounds will produce a basic aqueous solution at 25C? C) NaF F- is the conjugate base of the weak acid HF. The equilibrium for F- is: F- (aq) + H2O (l) ! HF (aq) + OH- (aq) So the solution will be basic. 2. Which of the following aqueous solutions will have the smallest percent ionization at 25C if the Ka of formic acid is 1.81 x 10-4 and the Ka of lactic acid is 1.38 x 10-4? B) 0.200 M lactic acid Smaller Ka means acid will dissociate less so must be lactic acid. Higher concentration will dissociate less, so choose 0.200 M lactic acid. 3. Which of the following solutions will have the highest pH? C) 0.10 M CH3NH2 (Kb = 4.2 x 104) All solutions have the same acid concentration, so can be directly compared by K values. Must convert to Ka using Ka = Kw/Kb. pH is directly correlated to Ka, the smallest Ka indicates the least dissociation or formation of H+ ions. 4. Which of the following solutions will be the best conductor of electrical current? C) potassium chloride, KCl (aq) Only electrolytes will conduct electricity, and strong electrolytes will conduct better than weak electrolytes. KCl is the only strong electrolyte. 5. For a neutral solution, it must be true that __________. [H+] = [OH-] 6. An aqueous solution of silver nitrate is added to an aqueous solution of potassium chromate, and this reaction produces a solid. What is the formula for the solid? Two possible products are potassium nitrate or silver chromate, nitrates are soluble, so the precipitate must be silver chromate: Ag2CrO4 7. For Mg(OH)2 (s), Ksp = 1.2 x 10-11. As the pH is raised, the solubility of Mg(OH)2 in water should __________. Increasing pH means decreasing the [H+], so the OH- ions will remain in solution (can't react to form water), which will shift the equilibrium to the left (reactants) and the solubility will decrease. 8. Methyl orange is an indicator with a Ka of 1 104. Its acid form, HIn, is red, while its base form, In, is yellow. At pH 6.0, the indicator will be _________. HIn (aq) ! H+ (aq) + In- (aq) At pH > pKa, the [H+] is less than the equilibrium concentration of the indicator, and the equilibrium of the indicator dissociation reaction will shift towards the products to form H+ and In-. Therefore, the solution will be yellow. 9. Addition of NH4Cl to an aqueous solution of NH3 will cause _______________. NH3 (aq) + H2O (l) ! NH4+ (aq) + OH- (aq) Adding NH4+ will increase [NH4+] and cause the equilibrium to shift towards the left (reactants). Thus [NH3] will increase, [OH-] will decrease and pH will decrease. 10. Arranging the following solutions, 0.20 M HF (Ka = 7.2 x 104), 0.20 M HOBr (Ka = 2.5 x 109), 0.20 M HClO2 (Ka = 1.1 x 102), in order of increasing pH would give _______________. Stronger acids have larger Ka and lower pH (more H+ dissociates): HClO2 < HF < HOBr 11. In comparing the three ionization constants for citric acid, H3C6H7O7, one would expect __________________. The Ka decreases as each proton dissociates from the acid. The equilibrium of the dissociation of the first proton is represented by Ka1, and so on. Ka1 (H3C6H7O7) > Ka2 (H2C6H7O7-) > Ka3 (HC6H7O72-) 12. Which of the following solutions will be the best buffer at a pH of 4.74? (Ka for HC2H3O2 is 1.8 x 10-5; Kb for NH3 is 1.8 x 10-5) Best buffer will have a pKa close to the desired pH and equal concentrations of acid and conjugate base. Thus we would mix equal molar solutions of HC2H3O2 and NaC2H3O2. 13. Hydroxylamine, HONH2, has a Kb of X. The Ka for HONH3+ would therefore be __________. Ka = Kw / X SHORT CALCULATION. 5 pts each 1. A mixture of HBrO (Ka = 2.5 x 109) and NaBrO (Kb = 4.0 x 106) would function best as a buffer at a pH of ________________. Best buffer has pKa = pH, so this would be best at: pK a = ! log 2.5 " 10 !9 = 8.60 = pH ( ) 2. Calculate the pH of a 0.1 M NaCN solution (Ka for HCN = 4.9 x 10-10) and choose the pH range that includes your answer. NaCN pH is determined by the equilibrium: CN- (aq) + H2O (l) ! HCN (aq) + OH- (aq) Kb = "CN ! $ # % "14 1 ! 10 Kb = = 2.0 ! 10 "5 4.9 ! 10 "10 ( x ) ( x ) # x 2 $ x = %OH " ' = 1.4 ! 10 "3 K b = 2.0 ! 10 "5 = & ( 0.1 " x 0.1 pOH = ! log "OH ! $ = ! log 1.4 & 10 !3 = 2.85 # % [ HCN ] "OH ! $ # % pH = 14 ! pOH = 14 ! 2.85 = 11.15 ( ) ( ) 3. When the equation Zn + As2O3 AsH3 + Zn2+ (acidic solution) is balanced using the smallest whole-number coefficients, what is the coefficient of H+? oxidation: reduction: (Zn Zn2+ + 2e-) x 6 As2O3 + 12H= +12 e- 2AsH3 + 3H2O overall rxn: 6Zn + As2O3 +12H= 6Zn2+ + 2AsH3 + 3H2O Coefficient on H+ is 12 4. A mixture of 82.49 g of aluminum and 117.65 g of oxygen is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete. 4 Al (s) + 3 O2 (aq) 2 Al2O3 (s) 1mol Al 1 equivalent = 3.06mol Al ! = 0.765 equivalents Al 26.98g Al 4mol Al 1mol O2 1 equivalent 117.65g O2 ! = 3.68mol O2 ! = 1.23 equivalents O2 32.00g O2 3mol O2 Al is limiting reactant 3mol O2 & # O2 remaining = 3.68mol O2 ! % 3.06mol Al " ( = 3.68mol O2 ! 2.30mol O2 = 1.38mol O2 $ 4mol Al ' 82.49g Al ! 1.38mol O2 ! 32.00g O2 = 44.2g O2 1mol O2 5. At a certain temperature, K for the reaction 2NO2 ! N2O4 -1 is 7.5 M . If 2.0 mol of NO2 is placed in a 2.0-liter container and permitted to react at this temperature, calculate the concentration of N2O4 at equilibrium. [ NO ] = 2 2.0mol NO2 = 1.0M NO2 2.0L Initial Change Equilibrium [NO2] 1.0 -2x 1.0 - 2x [N2O4] 0 +x x K = 7.5 = [N O ] = x [ NO ] (1 ! 2x ) 2 4 2 2 2 " 7.5x 2 ! 31x + 7.5 = 0 " x = [ N 2O4 ] = 0.39M 6. Consider the weak base hydroxylamine (HONH2), for which the Kb = 1.1 x10-8 at 25C. If 100.0 mL of a 0.500 M aqueous hydroxylamine solution is mixed with 100.0 mL of 0.500 M aqueous hydrochloric acid at 25C, the resulting solution will have pH 0.500mmol = 50.0mmol HONH 2 1mL 0.500mmol 100mL ! = 50.0mmol HCl 1mL Equal moles of acid and base mean equivalence point, pH will be determined by conjugate acid of hydroxylamine equilibrium. 100mL ! + ! HONH 3 # = " $ + 50.0mmol HONH 3 + = 0.25M HONH 3 200.0mL HONH3+ (aq) ! HONH2 (aq) + H+ (aq) Ka [ HONH ] ! H " = 2 + + ! HONH 3 # " $ # $ Initial Change Equilibrium [HONH3+] 0.25 -x 0.25-x [HONH2] 0 +x x [H+] 0 +x x Ka = K w 1 ! 10 "14 = = 9.1 ! 10 "7 "8 K b 1.1 ! 10 K a = 9.1 ! 10 "7 = x2 x2 # $ x = % H + ' = 4.8 ! 10 "4 M & ( 0.25 " x 0.25 pH = ! log " H + $ = ! log 4.8 & 10 !4 = 3.32 # % ( ) ( ) 7. Copper(II) oxalate is only slightly soluble in water, having Ksp = 2.9 x 10-8 at 25C. This means that the solubility of CuC2O4 in water at 25C is. CuC2O4 (s) ! Cu2+ (aq) + C2O42- (aq) 2 2 K sp = 2.9 ! 10 "8 = #Cu 2 + % #C2O4 " % = x 2 ' x = #Cu 2 + % = #C2O4 " % = 1.7 ! 10 "4 M $ &$ & $ & $ & 8. What volume of 0.235 M NaOH must be added to 60.00 mL of 0.200 M acetic acid (Ka = 1.76 x 10-5) to give a solution having a pH of 4.65? ! mol C2 H 3O2 = mol NaOH added ! mol HC2 H 3O2 initial = mol HC2 H 3O2 final + mol C2 H 3O2 final ! & "C2 H 3O2 $ ) ) 0.235M , ( x ) ( 0.06 + x ) L # % = ! log 1.76 , 10 !5 + log & pH = pK a + log ( + ( 0.200M , 0.06L ! 0.235M , x ( )) ( 0.06 + x ) L + '( * ' [ HC2 H 3O2 ] * ( ) 0.00948mol = 0.421M ! ( x ) " x = volume NaOH = 2.25 ! 10 #2 L x = volume NaOH = 0.041L # & 0.235M ! ( x ) ( 0.06 + x ) L 4.65 = 4.75 + log % ( $ ( 0.200M ! 0.06L " 0.235M ! ( x )) ( 0.06 + x ) L ' 0.235M ! ( x ) 0.79 = 0.012mol HC2 H 3O2 " 0.235M ! ( x ) 9. A monoprotic weak acid, when dissolved in water, is 0.92% dissociated and produces a solution with pH = 3.42. Calculate Ka for the acid. pH = 3.42 = ! log " H + $ & " H + $ = 3.8 ' 10 !4 M # % # % "4 3.8 ! 10 = 0.0092 ! [ HA ] # [ HA ] = 4.13 ! 10 "2 M ! H + # ! A% # 3.8 & 10 %4 Ka = " $ " $ = = 3.5 & 10 %6 M %2 4.13 & 10 [ HA ] 10. Calculate the pH of a solution prepared by mixing 50 mL of a 0.10 M solution of HF with 25 mL of a 0.20 M solution of NaF. pKa is 3.14. ( ) ( ) 2 0.20M & 25mL !2 "F! $ = # % ( 25mL + 50mL ) = 6.67 & 10 M [ HF ] = 0.10M ! 50mL = 6.67 ! 10 "2 M ( 25mL + 50mL ) & "F! $ ) & 6.67 , 10 !2 M ) pH = pK a + log ( # % + = 3.14 + log ( = 3.14 ' 6.67 , 10 !2 M + * ' [ HF ] * 11. You have solutions of 0.200 M HNO2 and 0.200 M KNO2 with which to prepare a buffer that has a pH of 3.000. What volumes of HNO2 and KNO2 are required to make 1 L of buffered solution? (Ka for HNO2 = 4.00 x 10-4) ! & " NO3 $ ) # % = ! log 4.00 , 10 !4 + log & 0.200M , ( x ) 1L ) pH = pK a + log ( + ( 0.200M , (1 ! x ) 1L + ' * ' [ HNO3 ] * ( ) # 0.200M ! ( x ) 1L & ( x) 3.000 = 3.40 + log % ) 0.398 = (1 " x ) $ 0.200M ! (1 " x ) 1L ( ' x = volume KNO3 = 0.285L ! volume HNO3 = 1L " 0.285L = 0.715L 12. A 50.00-mL sample of 0.100 M KOH is titrated with 0.100 M HNO3. Calculate the pH of the solution after 52.00 mL of HNO3 is added. Initial (mmol) Final (mmol) KOH 5 0 HNO3 5.2 0.2 H2O 0 5 pH determined by excess H+ in solution from complete dissociation of HNO3. 0.200mmol H + + !H # = = 1.96 % 10 &3 M " $ 102.00mL pH = ! log " H + $ = ! log 1.96 & 10 !3 = 2.708 # % ( ) ( ) 13. The solubility of Cd(OH)2 in water is 1.7 105 mol/L at 25C. What is Ksp for Cd(OH)2? Cd(OH)2 (s) ! Cd2+ (aq) + 2OH- (aq) K sp = !Cd 2 + # !OH % # = 1.7 & 10 %5 2 1.7 & 10 %5 " $" $ 2 ( )( ( )) 2 = 2.0 & 10 %14 14. Calculate the pH of a solution made by mixing equal volumes of a solution of HCl with a pH of 1.44 and a solution of HNO3 with a pH of 2.74. (Assume the volumes are additive.) HCl: pH = 1.44 = ! log " H + $ & " H + $ = 3.6 ' 10 !2 M # % # % HNO3: pH = 2.74 = ! log " H + $ & " H + $ = 1.8 ' 10 !3 M # % # % ( 2x new pH = ! log " H $ = ! log 1.9 & 10 !2 = 1.72 # % new ! H $ " + ( 3.6 % 10 ) x + (1.8 % 10 ) x = 1.9 % 10 #= &2 &3 ( ) ) &2 ( + ) ( ) 15. If 30 mL of 5.0 104 M Ca(NO3)2 is added to 70 mL of 2.0 104 M NaF, will a precipitate form? (Ksp of CaF2 = 4.0 1011) CaF2 (s) ! Ca2+ (aq) + 2F- (aq) K sp = !Ca 2 + # ! F % # " $" $ 5.0 % 10 &4 M ( 30mL ) 2+ !Ca # = = 1.5 % 10 &4 M " $ ( 30mL + 70mL ) 2 ( ) "F! $ = # % ( 2.0 & 10 !4 Q = 1.5 ! 10 "4 1.4 ! 10 "4 ( ( 30mL + 70mL ) M ( 70mL ) ) = 1.4 & 10 !4 M )( ) 2 = 2.9 ! 10 "12 Q < Ksp no precipitate will form 16. The mass percent of iron in an iron oxide is 77.7%. Find the empirical formula. 1mol Fe = 1.39mol Fe 55.85g Fe 1mol O 22.3g O ! = 1.39mol O 16.00g O 77.7g Fe ! 1:1 ratio of Fe:O empirical formula FeO 17. A 34.00 mL sample of 0.0291 M Ca(OH)2 required 24.75 mL of aqueous HCl for neutralization according to the reaction below. What is the concentration of the HCl Ca(OH)2 (aq) + 2 HCl (aq) ! CaCl2 (aq) + H2O (l) 34.00mL ! 0.0291M Ca (OH )2 = 0.989mmol Ca (OH )2 0.989mmol Ca (OH )2 ! 2mmol HCl = 1.98mmol HCl 1mmol Ca (OH )2 [ HCl ] = 1.98mmol HCl = 0.0800M HCl 24.75mL 18. The density of a certain diatomic gas is 1.696 g/L at STP. Identify the gas. m RT ! m $ PV = # & RT ' MM = ( " MM % V P L ! atm % " $ 0.08206 ' ( 273K ) # mol ! K & MM = 1.696 g L ! = 37.99 g mol (1atm ) 38 g mol gas is diatomic, so = 19 g mol gas is F2 2 19. Calculate the mass in grams of 3.65 x 1020 molecules of SO3. 3.65 ! 10 20 molecules ! 1mol 80g SO3 ! = 4.85 ! 10 "2 g 23 6.02 ! 10 molecules 1mol SO3 ...
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