mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 1 This print-out should have 45 questions. 6. The term is misleading, because the am- Multiple-choice questions may continue on monium ion is not an acid. the next column or page – find all choices Explanation: before answering. 003 10.0 points 001 10.0 points −9 If the value of Kb for pyridine is 1.8 × 10 , Assume that five weak acids, identified only calculate the equilibrium constant for by numbers (1, 2, 3, 4, and 5), have the + C5H5NH (aq)+ H2O(ℓ) → following ionization constants. + C5H5N(aq) + H3O (aq) . − Ionization 1. 5.6 × 10 6 correct Acid Constant − Ka value 2. 1.8 × 10 9 − 1 1.0 × 10 3 − × −16 2 3.0 × 10 5 3. 1.8 10 − 3 2.6 × 10 7 − × 8 4 4.0 × 10 9 4. 5.6 10 −11 5 7.3 × 10 − 5. −1.8 × 10 9 The anion of which acid is the strongest Explanation: base? 004 10.0 points 1. 4 Which of the following is true in pure water at any temperature? 2. 5 correct 1. Kw decreases with increasing tempera- 3. 2 ture. + − −14 4. 3 2. [H3O ][OH ] = 1.0 × 10 + − 5. 1 3. [H3O ] = [OH ] correct Explanation: 4. pH = 7.0 or greater than 7.0 002 10.0 points 5. pH = 7.0 The term “Ka for the ammonium ion” de- scribes the equilibrium constant for which of Explanation: the following reactions? Kw is shown to INCREASE with increas- ing temperature. pH = 7 is only true when + ⇀ + ◦ + − 1. NH4 +H2O ↽ NH3 +H3O correct water is at 24 C. [H3O ][OH ] = Kw, which increases with temperature. + − 2. NH3 +H2O ↽⇀ NH4 + OH At high temperatures pH can be less than + − 7. Thus [H3O ] = [OH ] is the only case that + ⇀ + 3. NH3 +H3O ↽ NH4 +H2O is true. + − ⇀ 4. NH4 + OH ↽ NH3 +H2O 005 10.0 points Which is NOT a conjugate acid-base pair? + − 5. NH4Cl(solid) + H2O ↽⇀ NH + Cl 4 − 1. H2O:OH mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 2 + Kw − [H O ] = 2. HCl : Cl 3 [OH−] 1.0 × 1014 + × −6 3. H SO : H SO = − =3.0 10 M 3 4 2 4 3.3 × 10 9 − 4. H : H 2 008 10.0 points − 2− What is [OH ] in a 0.0050 M HCl solution? 5. H2SO4 : SO4 correct − Explanation: 1. 6.6 × 10 5 M 2− Except for H2SO4 and SO4 , the members − of all of the pairs differ by one proton. 2. 5.0 × 10 3 M − 006 10.0 points 3. 1.0 × 10 7 M − What is the conjugate acid of NO ? 3 − 4. 2.0 × 10 12 M correct − 1. NO2 5. 1.0 M 2. NH3 Explanation: − 3. H+ [OH ]=0.0050 M Since HCl is a strong acid, it completely dissociates and H+ is 0.0050 M. 4. HNO3 correct − 2− HCl ↽⇀ H+ + Cl 5. NO3 − 6. OH + − × −14 Explanation: Kw = [H ][OH ]=1 10 − K Since the question asks for the conjugate [OH ] = w − [H+] acid, we can assume NO3 is acting as a base. −14 This means that it is a proton acceptor. To 1 × 10 − = =2 × 10 12 M form the conjugate acid, it accepts a H making 0.0050 HNO3. 007 10.0 points 009 10.0 points + − × −9 Which pH represents a solution with 1000 What is [H3O ] when [OH ]=3.3 10 M? − times higher [OH ] than a solution with pH − 1. 1.0 × 10 7 M of 5? − 2. 3.3 × 10 9 M 1. pH=2 − 3. 3.3 × 10 5 M 2. pH = 0.005 − 4. 3.0 × 10 6 M correct 3. pH=8 correct − 5. 6.6 × 10 5 M 4. pH=1 Explanation: − − 5. pH=3 [OH ]=3.3 × 10 9 M + − 14 Kw = [H3O ][OH ]=1 × 10 6. pH = 5000 mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 3 Hydroxylamine is a weak molecular base with −9 7. pH=7 Kb =6.6 × 10 . What is the pH of a 0.0500 M solution of hydroxylamine? 8. pH=4 1. pH = 8.93 9. pH=6 Explanation: 2. pH = 7.12 pH=5 3. pH = 3.63 pOH = 14 − pH = 14 − 5=9 4. pH = 4.74 − − − [OH ]=10 pOH = 10 9 M 5. pH = 9.26 correct − − 3 −9 [OH ]x = 1000 [OH ] = (10 )(10 M) − 6. pH = 9.48 = 10 6 M 7. pH = 10.37 pOHx = − log(OHx)=6 pHx = 14 − pOHx = 14 − 6=8 Explanation: Hydroxylamine is a weak base, so use the − equation to calculate weak base [OH ] con- 010 10.0 points centration (note that this is the approximate What isthe pH of a 0.12 M Ba(OH)2 aqueous equation. Why? Because Kb is very small solution? and the concentration is reasonable) : − [OH ] = K C 1. 1.33802 b b = p(6.6 × 10−9)(0.0500) 2. 8.7 − =1q.82 × 10 5 − 3. 0.619789 After finding [OH ], you can find pH using either method below: A) 4. 13.3802 correct − pOH = − log 1.82 × 10 5 =4.74 5. 10.0352 − pH = 14 4.74=9.26 Explanation: or B) [Ba(OH) ]=0.15 M + Kw 2 [H ] = − Ba(OH)2 is a strong base which dissociates [OH ] × −14 in aqueous solution to produce two moles of 1.0 10 × −10 − = − =5.52 10 OH for every mole of Ba(OH)2, so 0.12 M 1.82 × 10 5 − − Ba(OH)2 produces 0.24 M OH . pH = − log 5.52 × 10 10 =9.26 2+ − Ba(OH)2 → Ba + 2OH ini 0.12M 0M 0M 012 10.0 points ∆ −0.12M +0.12M 2(0.12 M) What is the pH of a 0.2 M solution of potas- sium generate (KR-COO)? Ka for the generic fin 0 M +0.12M +0.24 M − acid (R-COOH) is 2.7 × 10 8. pH = 14−pOH = 14−(− log0.24) = 13.3802 1. 10.285 2. 7.000 011 10.0 points mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 4 3. 10.565 Your answer must be within ± 0.4% Correct answer: 2.70956. 4. 10.195 Explanation: − M =0.16 M K =4.2 × 10 10 5. 3.565 C6H5NH3NO3 b It’s a salt of a weak base (BHX).This means you need a K for the weak acid BH+: 6. 7.569 a Kw 7. 6.431 Ka = Kb 1.0 × 10−14 8. 3.435 = 4.2 × 10−10 −5 9. 10.435 correct =2.38095 × 10 10. 10.805 You CAN use the approximation for the equilibrium which means that Explanation: −8 MKR−COO =0.2 M Ka =2.7 × 10 + [H ] = Ka · CBH+ It’s a salt of a weak generic acid (KA). Get it? Generic acid makes generic ions. = p(2.38095 × 10−5)(0.16) Ha! This means you need a Kb for the weak =0q.0019518 M − Kw base A . Use Kb = and you’ll get the Ka pH = − log(0.0019518) = 2.70956 −7 Kb = 3.7037 × 10 . You CAN use the ap- proximation for the equilibrium which means that 015 10.0 points − The pH of lemon juice is approximately 2.4. · − [OH ] = Kb CA =0.000272166 M At this pH, the hydronium ion concentration p is closest to which value? pH = 14 − pOH −12 = 14+ log(0.000272166) = 10.4348 1. 2.50 × 10 M − 2. 5.62 × 10 4 M 013 10.0 points ◦ − At 25 C, the pH of a water solution of a salt 3. 4.00 × 10 3 M correct of a WEAK acid and a STRONG base is 4. 250 M 1. less than 7. Explanation: pH = 2.4, so 2. greater than 7. correct −2.4 M + = 10 =0.00398107 M 3. about 7. H 4. equal to the hydrogen ion concentration. 016 10.0 points Explanation: Which solution has the highest pH? 014 10.0 points 1. 0.1 M of KHCOO, −4 What is the pH of a 0.16 M solution of anilin- Ka HCOOH =1.8 × 10 ium nitrate (C6H5NH3NO3)? Kb for aniline −10 is 4.2 × 10 . 2. 0.1 M of KCl, Ka HCl = very large mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 5 3. 0.1 M of KCH3COO, 019 10.0 points × −5 Ka HC2H3O2 =1.8 10 A solution has a pH of 4.35. Find the pOH. × −4 4. 0.1 M of KNO2, Ka HNO2 =4.5 10 1. 4.35 −8 5. 0.1 M of KClO, Ka HClO = 3.5 × 10 2. 9.65 correct correct 3. None of these Explanation: 4. 18.35 017 10.0 points What is the pH of a solution that contains Explanation: 11.7 g of NaCl for every 200 mL of solution? pH = 4.35 1. 1.0 pOH = 14 − pH=9.65 − 2. 10 1 020 (part 1 of 2) 10.0 points 3. 7.0 correct The pH of an aqueous solution is measured as + 1.21. Calculate the [H3O ]. − 4. 1.0 × 10 7 Correct answer: 0.0616595 M. Explanation: m =11.7 g V = 200 mL Explanation: NaCl soln + NaCl completely dissociates in water to give pH=1.21 [H3O ] = ? − Na+ and Cl , neither of which hydrolyzes and + − pH = − log[H O+] so in aqueous NaCl the H3O and OH ions 3 + result from autoionization of water. log[H3O ] = −pH + [H3O ] = antilog(−pH) + − −14 × −pH Kw = [H3O ][OH ]=1 × 10 =1 10 + − −7 × −1.21 [H3O ] = [OH ]=1 × 10 =1 10 =0.0616595 M + and pH = − log[H3O ]=7.0 021 (part 2 of 2) 10.0 points − 018 10.0 points Calculate the [OH ].