Chemistry 192 Problem Set 4 Spring, 2018 Solutions

Transcription

1 Chemistry 192 Problem Set 4 Spring, 2018 Solutions 1. The ionization constant of benzoic acid in water associated with the reaction C 6 H 5 COOH (aq) + H 2 O (l) C 6 H 5 COO (aq) + H 3O + (aq) is K a = Calculate the value of K b associated with the reaction C 6 H 5 COO (aq) + H 2O (l) C 6 H 5 COOH (aq) + OH (aq). K a K b = [H 3O + ][C 6 H 5 COO ] [C 6 H 5 COOH][OH ] [C 6 H 5 COOH] [C 6 H 5 COO ] K b = K b = = K w 2. Use the data and/or results from problem 1 to calculate the ph of a M solution of sodium benzoate, NaC 6 H 5 COO. C 6 H 5 COO (aq) + H 2O (l) C 6 H 5 COOH (aq) + OH (aq) [C 6 H 5 COO ] [C 6 H 5 COOH] [OH ] initial M 0 M 0 M change -x x x equilibrium ( x) M x M x M K b = x 2 = x x = x = [OH ] = poh = log 10 [OH ] = 5.40 ph = poh =

2 3. The ionization constant of hydroxylamine producing the hydroxylammonium ion via the reaction HONH 2(aq) + H 2 O (l) HONH + 3(aq) + OH (aq) is K b = Calculate the ph of a M solution of hydroxylammonium chloride. HONH + 3(aq) + H 2O (l) H 3 O + (aq) + HONH 2(aq) K a = K w K b = = [HONH + 3 ] [HONH 2 ] [H 3 O + ] initial M 0 M 0 M change -x x x equilibrium ( x) M x M x M x x = x2 = x = [H 3 O + ] = = ph = log 10 [H 3 O + ] = Using the ionization constant of acetic acid CH 3 COOH (aq) + H 2 O (l) CH 3 COO (aq) + H 3O + (aq) K a = calculate the mass of sodium acetate that must be added to 100. produce a solution with ph=8.00. CH 3 COO (aq) + H 2O (l) CH 3 COOH (aq) + OH (aq) ml of water to K b = K w K a = = [OH ] = [CH 3 COOH] = = ( ) 2 [CH 3 COO ] = [CH 3 COO ] = M n sodium acetate = ( mol L 1 )(0.100 L) = mol m = (82. g mol 1 )( mol) = g 2

3 5. The ionization reaction of hypochlorous acid with water HOCl (aq) + H 2 O (l) OCl (aq) + H 3O + (aq) has a pk a = Consider a M NaOCl (sodium hypochlorite) solution. (a) Do you expect the ph of the NaOCl solution to be less than or greater than 7.00? Explain your reasoning. NaOCl is the salt of a weak acid and a strong base. The ph should be greater than (b) Calculate the ph of a M NaOCl solution given sodium hypochlorite is completely ionized in water. OCl (aq) + H 2O (l) HOCl (aq) + OH (aq) K b = [HOCl][OH ] [OCl ] = 10 ( ) = [OCl ] [HOCl] [OH ] initial M 0 M 0 M change -x x x equilibrium ( x) M x M x M x x x = x = [OH ] = = M poh = log 10 ( ) = 3.73 ph = poh = The ionization constant for hydrofluoric acid via the reaction HF (aq) + H 2 O (l) H 3 O + (aq) + F (aq) is K a = Calculate the concentration of fluoride ion in a M hydrofluoric acid solution and an aqueous mixture M aqueous HF and M HCl. Recall that HCl is completely ionized in water. [H 3 O + ][F ] = [HF] Without HCl [HF] [F ] [H 3 O + ] initial M 0 M 0 M change -x x x equilibrium ( x) M x M x M 3

4 x = ± x x = x x = 0 where only the positive solution is physical. With HCl ( ) 2 + 4( ) 2 [HF] [F ] [H 3 O + ] initial M 0 M M change -x x x equilibrium ( x) M x M ( x) M = x( x) x = x x = 0 x = ± ( ) 2 + 4( ) 2 7. Aqueous methylamine is a weak base with the reaction = CH 3 NH 2(aq) + H 2 O (l) OH (aq) + CH 3NH + 3(aq) having ionization constant K b = A M solution of mehtylamine is mixed with sufficient sodium hydroxide to yield a solution having ph=9.30. Calculate the concentration of the methylammonium ions and methylamine in the solution. [OH ][CH 3 NH + 3 ] = [CH 3 NH 2 ] [OH ] = 10. ( ) = [CH 3 NH 2 ] [CH 3 NH + 3 ] [OH ] initial M 0 M 0 M change -x x M equilibrium ( x) M x M M 4

5 x( ) x = x = x x = [CH 3 NH + 3 ] = M [CH 3 NH 2 ] = x = M Note: The Hendersen-Hasselbalch equation would fail for this problem, because the full quadratic equation is required for the solution. 8. The value of K a for the dissociation of benzoic acid C 6 H 5 COOH (aq) + H 2 O (l) H 3 O + (aq) + C 6H 5 COO (aq) is Calculate the ph of a solution that is M in benzoic acid and M in sodium benzoate. We solve this problem two ways. First, we apply the Hendersen-Hasselbalch equation ph = pk a + log 10 [C 6 H 5 COO ] [C 6 H 5 COOH] [C 6 H 5 COO ][H 3 O + ] [C 6 H 5 COOH] = K a = pk a = log 10 ( ) = 4.20 ph = log = 4.20 Alternatively, we can solve the problem with an ICE table: [C 6 H 5 COOH] [C 6 H 5 COO ] [H 3 O + ] initial M M 0 M change -x x x equilibrium ( x) M x M x M x( x) x x = ph = log =

6 9. For benzoic acid as in problem 8, calculate the ph of a solution that is M in benzoic acid and M in sodium benzoate. Again, we can either use the Hendersen-Hasselbalch equation or an ICE table. Using the Hendersen-Hasselbalch equation first Using an ICE table ph = log = 4.87 [C 6 H 5 COOH] [C 6 H 5 COO ] [H 3 O + ] initial M M 0 M change -x x x equilibrium x M x M x M x( x) = x x = [H 3 O + ] ( ).01 = = ph = log = Sodium azide (NaN 3 ) is a highly toxic and highly water soluble salt made from sodium hydroxide (NaOH, a strong base) and hydrazoic acid (HN 3, a weak acid). Sodium azide solutions are completely ionized in water. It is found that a M aqueous solution of sodium azide has a ph=8.80. Calculate the pk a of the reaction HN 3(aq) + H 2 O (l) N 3(aq) + H 3O + (aq). N 3(aq) + H 2O (l) HN 3(aq) + OH (aq) poh = = 5.20 [OH ] = = [N 3 ] [HN 3 ] [OH ] initial M 0 M 0 M change equilibrium M M M K b = [HN 3][OH ] [N 3 ] = ( ) = K a = K w K b = = pk a = log 10 ( ) =

7 11. For aqueous ammonia using the reaction NH 3(aq) + H 2 O (l) NH + 4(aq) + OH (aq) K b = calculate the ph of a solution that is M in ammonia and M in ammonium chloride. Once again, we have the choice of solving this problem using the Hendersen- Hasselbalch equation or an ICE table. For the Hendersen-Hasselbalch equation, NH + 4(aq) + H 2O (l) NH 3(aq) + H 3 O + (aq) K a = K w K b = = pk a = log 10 ( ) = 9.26 ph = pk a + log 10 [NH 3 ] [NH + 4 ] = log = 8.56 Using an ICE table [NH + 4 ][OH ] [NH 3 ] = [NH 3 ] [NH + 4 ]] [OH ] initial M M 0 M change -x x x equilibrium ( x) M x M x M [OH ] = x( x).0100 x = ( ) = poh = log = 5.44 ph = poh = Calculate the ph when 100. ml of the solution discussed in problem 8 is combined with 20.0 ml of M hydrochloric acid. From problem 8 before the addition of the hydrochloric acid, the concentrations of all species are [H 3 O + ] = [C 6 H 5 COO ] = [C 6 H 5 COOH] M 7

8 The initial number of moles of benzoic acid and the benzoate ion are n C6 H 5 COOH = n C6 H 5 COO = ( mol L 1 )(0.100 L) = mol The number of added moles of hydronium ion is given by n H3 O + = ( mol L 1 )( L) = mol From the reaction given in Problem 8 and LeChâtelier s principle, the addition of acid shifts the equilibrium to the left. Then C 6 H 5 COOH C 6 H 5 COO H 3 O + initial mol mol 0 mol change mol mol x equilibrium mol mol x mol Letting y = [H 3 O + ] ( ) mol y L ( ) = mol L y = [H 3 O + ] = M ph = Calculate the ph when 100. ml of the solution discussed in problem 11 is combined with 20.0 ml of M hydrochloric acid. Initially n NH + 4 = ( mol L 1 )(0.100 L) = mol n NH3 = ( mol L 1 )(0.100 L) = mol The amount of added hydronium ion is n H3 O + = ( L)( mol L 1 ) = mol From the reaction given in Problem 11 and LeChâtelier s principle, the addition of acid shifts the equilibrium to the right. Then NH 3 NH + 4 OH initial mol mol 0 mol change mol mol x mol equilibrium mol mol x mol 8

9 Let y = [OH ]. Then ( ) mol y L ( ) = mol L y = [OH ] = M poh = log 10 ( ) = 5.56 ph = poh = Calculate the ph when 100. ml of the solution discussed in problem 8 is combined with 20.0 ml of a M aqueous sodium hydroxide solution. Initially n C6 H 5 COOH = n C6 H 5 COO = ( mol L 1 )(0.100 L) = mol Hydroxide ion added n OH = ( mol L 1 )( L) = mol C 6 H 5 COOH C 6 H 5 COO H 3 O + initial mol mol 0 mol change mol mol x equilibrium mol mol x mol Let y = [H 3 O + ]. Then ( ) mol y L ( ) = mol L y = [H 3 O + ] = M ph = From the dissociation constant of benzoic acid given in problem 8, calculate the number of grams of sodium benzoate that must be added to 100. ml of M benzoic acid to produce a solution buffered to ph = ph = pk a + log 10 [C 6 H 5 COO ] [C 6 H 5 COOH] 4.00 = log 10 ( ) + log 10 [C 6 H 5 COO ]

10 [C 6 H 5 COO ] = M Assuming negligible volume change by the addition of the sodium benzoate number of moles = ( mol L 1 )(0.100 L) = mol mass = (144 g mol 1 )( mol) = 0.91 g 16. The weak base diethylamine reacts with water according to the reaction (C 2 H 5 ) 2 NH (aq) + H 2 O (l) (C 2 H 5 ) 2 NH + 2(aq) + OH (aq), and the equilibrium constant for the reaction is K b = A buffer solution is formed by mixing diethylamine and diethylammonium chloride having concentrations [(C 2 H 5 ) 2 NH]=0.124 M and [(C 2 H 5 ) 2 NH + 2 ]=0.224 M. Calculate the ph of the final solution when L of the buffer are mixed with L of M NaOH (a strong base). n (C2 H 5 ) 2 NH = (0.124 mol L 1 )(0.400 L) = mol n (C2 H 5 ) 2 NH + 2 = (0.224 mol L 1 )(0.400 L) = mol n OH = ( mol L 1 )( L) = mol Method 1: Hendersen-Hasselbalch: (C 2 H 5 ) 2 NH + 2(aq) + H 2O (l) (C 2 H 5 ) 2 NH (aq) + H 3 O + (aq) K a = K w K b = = pk a = log 10 ( ) = ph = pk a + log 10 [(C 2 H 5 ) 2 NH] [(C 2 H 5 ) 2 NH + 2 ] = log 10 [( )]/0.550 [( )]/0.550 = Method 2: ICE Table: n (C2 H 5 ) 2 NH n (C2 H 5 ) 2 NH + n OH initial mol mol 0 mol change mol M mol y mol equilibrium mol mol y mol 10

11 = [OH ](0.0873/0.550) (0.0519/0.550) [OH ] = M poh = log 10 ( ) = 3.39 ph = = Pyridine (C 5 H 5 N) reacts with water as a weak base according to the reaction C 5 H 5 N (aq) + H 2 O (l) C 5 H 5 NH + (aq) + HO (aq) with associate base ionization constant K b = A buffer is made by combining aqueous pyridine and pyridinium chloride (C 5 H 5 NHCl) of concentrations [C 5 H 5 N]=0.200 M and [C 5 H 5 NH + ]=0.300 M. A L sample of the buffer is then mixed with L of M aqueous hydrochloric acid (HCl, a strong acid). Calculate the ph of the mixture of the buffer and hydrochloric acid. Methd 1, Hendersen-Hasselbalch C 5 H 5 NH + (aq) + H 2O (l) C 5 H 5 N (aq) + H 3 O + (aq) K a = Before mixing = pk a = log 10 ( ) = 5.18 Method 2, ICE Table Before mixing n C5 H 5 N = (0.200 mol L 1 )(0.100 L) = mol n C5 H 5 NH + = (0.300 mol L 1 )(0.100 L) = mol n H3 O + = (0.235 mol L 1 )( L) = mol [C 5 H 5 N] ph = pk a + log 10 [C 5 H 5 NH + ] = log = n C5 H 5 N = (0.200 mol L 1 )(0.100 L) = mol n C5 H 5 NH + = (0.300 mol L 1 )(0.100 L) = mol n H3 O + = (0.235 mol L 1 )( L) = mol 11

12 n C5 H 5 N n C5 H 5 NH + n OH initial change y equilibrium y K b = [C 5H 5 NH + ][OH ] [C 5 H 5 N] = ( ) [OH 2 ] ( ) [OH ] = M poh = log 10 ( ) = 9.08 ph = = The base dissociation constant of phenylamine (C 6 H 5 NH 2 ) is K b = A buffer is prepared that is M in phenylamine and M in phenylammonium cation (C 6 H 5 NH + 3 ). A L sample of the buffer is then mixed with L of M sodium hydroxide (a stong base). Calculate a) the ph of the initial buffer solution, and 2) the ph of the buffer/sodium hydroxide mixture. Method 1 - ICE Table Buffer C 6 H 5 NH 2(aq) + H 2 O (l) C 6 H 5 NH + 3(aq) + OH (aq) [C 6 H 5 NH 2 ] [C 6 H 5 NH + 3 ] [OH ] initial M M 0 M change -y y y equilibrium ( y) M ( y) M y M = [C 6H 5 NH + 3 ][OH ] [C 6 H 5 NH 2 ] = y( y) y 2y y = [OH ] = M poh = log 10 ( ) = 9.60 ph = poh = 4.40 Mixture n C6 H 5 NH 2 = (0.100 mol L 1 )(0.100 L) = mol n C6 H 5 NH + 3 = (0.200 mol L 1 )(0.100 L) = mol n OH = ( mol L 1 )(0.100 L) = mol 12

13 n C6 H 5 NH 2 n C6 H 5 NH + 3 initial change y equilibrium y n OH = ( ) [OH 2 ] ( ) [OH ] = M poh = log 10 ( ) = 9.54 ph = poh = 4.46 Method 2 - Henderson-Hasselbalch Buffer Mixture K a = K w K b = C 6 H 5 NH + 3(aq) + H 2O (l) C 6 H 5 NH 2(aq) + H 3 O + (aq) = pk a = log 10 ( ) = 4.70 ph = pk a + log 10 [C 6 H 5 NH 2 ] [C 6 H 5 NH + 3 ] = log = ph = log = A buffer of volume L is M in formic acid (HCOOH) and M in the formate (HCOO ) anion. The buffer is then combined with L of M sodium hydroxide. Given the pk a of formic acid is 3.74, calculate the ph of the mixture of the buffer with the NaOH solution. Approximations work for this problem. HCOOH (aq) + H 2 O (l) HCOO (aq) + H 3O + (aq) n OH Method 1, Henderson-Hasselbalch n HCOOH = (0.250 mol L 1 )(0.400 L) = mol n HCOO = (0.350 mol L 1 )(0.400 L) = mol added = ( mol L 1 )(0.100 L) = mol ph = pk a + [HCOO ] [HCOOH] 13

14 Method 2, ICE table = = K a = = n HCOOH n HCOO n H3 O + initial mol mol 0 mol change mol mol y mol equilibrium mol mol y mol = [H 3O + ] [H 3 O + ] = M ph = log 10 ( ) = Ethanolamine, HO(CH 2 ) 2 NH 2 is a weak base in aqueous solutions, and its conjugate acid, HO(CH 2 ) 2 NH + 3 is called the ethanolammonium ion. The pk b of ethanolamine is A buffer is formed by mixing 0.10 moles of HO(CH 2 ) 2 NH 2 with 0.20 moles of HO(CH 2 ) 2 NH + 3 to make an aqueous solution having a final total volume of L. Calculate a) the ph of the buffer, and b) the ph if the buffer is mixed with L of M hydrochloric acid (a strong acid). Approximations work for this problem. Method I, Henderson-Hasselbalch HO(CH 2 ) 2 NH + 3(aq) + H 2O (l) HO(CH 2 ) 2 NH 2(aq) + H 3 O + (aq) Original Buffer After mixing pk a = pk b = 9.50 ph = pk a + log 10 [HO(CH 2 ) 2 NH 2 ] [HO(CH 2 ) 2 NH + 3 ] = log = 9.20 n H3 O + = (0.050 mol L 1 )(0.100 L) = mol ph = log =

15 Method II, ICE Table HO(CH 2 ) 2 NH 2(aq) + H 2 O (l) HO(CH 2 ) 2 NH + 3(aq) + OH (aq) Before Mixing K b = = [HO(CH 2 ) 2 NH 2 ] [HO(CH 2 ) 2 NH + 3 ] [OH ] initial 0.10/0.50 M 0.20/0.50 M 0 M change -y y y equilibrium 0.20-y M y M y M y( y) 0.20 y 0.40y 0.20 = After Mixing y = [OH ] = poh = log 10 ( ) = 4.80 ph = poh = 9.20 n HO(CH2 ) 2 NH 2 n HO(CH2 ) 2 NH + 3 initial change y equilibrium y n OH [OH ] = [OH ] = poh = 4.83 ph = Ethyl amine (C 2 H 5 NH 2 ) is a weak base with pk b = A buffer is made by combining moles of ethyl amine and moles of an ethyl ammonium salt (C 2 H 5 NH + 3 ) with water so that the total volume is 0.25 L. The buffer is then combined with L of M hydrochloric acid (HCl). Calculate 1) the initial ph of the buffer, and 2) the ph of the buffer after the addition of the hydrochloric acid. Approximations work for this problem. Method 1, Henderson-Hasselbalch C 2 H 5 NH + 3(aq) + H 2O (l) C 2 H 5 NH 2(aq) + H 3 O + (aq) 0 Name: 15

16 Before the addition of HCl After the addition of HCl Method 2, ICE table pk a = pk b = ph = pk a + log 10 [C 2 H 5 NH 2 ] [C 2 H 5 NH + 3 ] ph = log / /0.25 = n H3 O + = (0.100 mol L 1 )(0.020 L) = mol ph = log 10 ( )/0.27 ( )/0.27 = C 2 H 5 NH 2(aq) + H 2 O (l) C 2 H 5 NH + 3(aq) + OH (aq) Before mixing K b = = [C 2 H 5 NH 2 ] [C 2 H 5 NH + 3 ] [OH ] initial 0.025/0.25 M 0.050/0.25 M 0 M change -y y y equilibrium 0.10-y M y M y M = y( y) 0.10 y y(0.20) 0.10 y = [OH ] = After mixing poh = log 10 ( ) = 3.68 ph = poh = n H3 O + = (0.100 mol L 1 )(0.020 L) = mol n C2 H 5 NH 2 n C2 H 5 NH + 3 initial M change y equilibrium y n OH = [OH ](0.052) [OH ] = M poh = log 10 ( ) = 3.72 ph = poh =