CHEMISTRY 1AA3 Tutorial 2 Answers - WEEK E WEEK OF JANUARY 22, 2001 M.A. Brook B.E. McCarry A. Perrott 1. (i) What is the conjugate base of each of the following species? (a) H 3 O + (b) NH 4 + (c) HCl (d) H 3 PO 4 (a) water, H2O (b) ammonia, NH 3 (c) chloride ion, Cl (d) dihydrogen phosphate ion, H 2 PO 4 (ii) What is the conjugate acid of each of the following species? (a) O2 (b) PO4 3- (c) CO3 2- (d) HSO4 - (a) hydroxide ion, OH (b) hydrogen phosphate ion, 2 HPO 4 (c) hydrogen carbonate ion, HCO 3 (d) sulfuric acid H 2 SO 4 2. (a) What is the ph of a 0.04M solution of Ba(OH 2 ) in water? (b) A 20.0 ml sample of a Ba(OH) 2 solution when diluted to 720 ml had a ph of 10.45. What was the concentration of Ba(OH) 2 in the original 20 ml solution? (a) Ba(OH) 2 Ba 2+ + 2OH - 0.04M 0.04M 0.08M [OH - ] = 0.08M and poh = 1.10
ph = 14 - poh = 14-1.10 = 12.90 (b) ph = 10.45 poh = 14-10.45 = 3.55 [OH] = antilog(-3.55) = 2.82 x 10-4 M [Ba(OH) 2 ] = 0.5 x (2.82 x 10-4 ) = 1.41 x 10-4 M in 720 ml solution [Ba(OH) 2 ] in 20 ml solution = 1.41 x 10-4 M x (720/20) = 5.08 x 10-3 M 3. Nitrous acid, HNO2, is a weak acid with an ionization constant of 4.5 x 10 4. (a) What is the equilibrium concentration of NO 2 in a 0.25 M aqueous solution of HNO 2? (b) What is the ph of a 0.25 M aqueous solution of HNO 2? (a) EQUATION: HNO 2 + H 2 O Ξ H 3 O + + NO 2 - EQUILIBRIUM EXPRESSION: QUANTITIES: Ka = [H3O + ] [NO2 - ] = 4.5 x 10-4 [HNO 2 ] HNO 2 + H 2 O Ξ H 3 O + + NO 2 - Initial: 0.25 0 0 Equilibrium: 0.25-x x x Thus, 4.5 x 10-4 = x. x 0.25-x Assume that x << 0.25 and solve problem then check later if assumption is valid; Equation becomes: 4.5 x 10-4 = x. x
0.25 and x = [NO 2 - ] = [H 3 O + ] = 1.06 x 10-2 M [Check assumption: 1.06 x 10-2 corresponds to: 1.06 x 10-2 x 100% = 4.2% 0.25 This number is < 5%; thus, assumption is valid to use here]. (b) ph = -log [1.06 x 10-2 ] = 1.97 4. Determine the [OH ], ph and the percent dissociation of a 0.50 M solution of dimethylamine, (CH 3 ) 2 NH. The K b of dimethylamine is 7.4 x 10 4. EQUATION: (CH 3 ) 2 NH + H 2 O Ξ (CH 3 ) 2 NH 2 + + OH - EQUILIBRIUM EXPRESSION: K b = [(CH 3 ) 2 NH 2 + ] [OH - ] = 7.4 x 10-4 [(CH 3 ) 2 NH] QUANTITIES: (CH3)2NH + H2O Ξ (CH3)2NH2 + + OH - Initial: 0.50 0 0 Equilibrium: 0.50-x x x THUS: 7.4 x 10-4 = x. x Assume x<<0.20 and equation becomes: 0.50-x x 2 = (0.50)(7.4 x 10-4 ) and x = [OH - ] = 1.92 x 10-2 M [Check assumption: 1.92 x 10-2 x 100% = 3.8%; x < 5%; thus, assumption is valid.]
0.50 poh = -log[oh - ] = -log(1.92 x 10-2 ) = 1.72 and ph = 14.0 - poh = 14.0-1.72 = 12.28 Percent dissociation = 1.92 x 10-2 x 100% = 3.8% 0.50 5. A student dissolved 2.15 g of hydrazoic acid, HN 3, to give one litre of solution. The ph of the solution was measured and found to be 3.01. What is the ionization constant of hydrazoic acid? EQUATION: HN 3 + H 2 O Ξ H 3 O + N 3 - EQUILIBRIUM EXPRESSION: K a = [H 3 O + ] [N 3 - ] [HN 3 ] QUANTITIES: Ka is unknown, but the ph of a 2.15g = 0.05M solution is 3.01. 43g/mol [H 3 O + ] = antilog(-ph) = antilog (-3.01) = 9.77 x 10-4 M and from the chemical equation, [H 3 O + ] = [N 3 - ] = 9.77 x 10-4 M and [HN 3 ] = 0.05 - [H 3 O + ] = 0.0490 M K a = (9.77 x 10-4 )( 9.77 x 10-4 ) = 1.95 x 10-5 0.049 6. (a) A 0.300 M solution of trimethylammonium chloride, N(CH 3 ) 3. HCl, has a ph of 5.20. Calculate the values of K b for trimethylamine, N(CH 3 ) 3. (b) What is the ph of 0.400 M (CH 3 ) 3 in water? (c) What is the ph of the solution resulting from the mixing of 200 ml of 0.300 M N(CH 3 ) 3 and 150 ml of 0.300 M N(CH 3 ) 3. HCl? (d) What is the ph of the solution in part (c) after dilution to 2.00 L in water? (a) N(CH3)3H + + H2O Ξ N(CH3)3 + H3O +
Initial 0.300 M 0 0 Change -x +x +x Equilibrium 0.300-x x x In this case we are given information about the equilibrium concentrations of these species, since we are told that ph = 5.20. This means that at equilibrium, [H 3 O + ] = antilog (ph) = 10 5.2 = 6.31 10 6 M This is the value for x in our table, and so we can write K a = [N(CH 3 ) 3 ][ H 3 O + ] = (6.31 10 6 ) 2 = 1.33 10 10 [N(CH3)3H + ] (0.300-6.31 10 6 ) K b = K w = 1.0 10 14 = 7.52 10 5 K a 1.33 10 10 (b) This is just a weak base problem: N(CH 3 ) 3 + H 2 O Ξ N(CH 3 ) 3 H + + OH Initial 0.400 M 0 0 Change -x +x +x Equilibrium 0.400-x x x K b = 7.52 10 5 = [N(CH 3 ) 3 H + ][ OH] = x 2 Assuming x << 0.400, this solves to give [N(CH3)3] x = 5.48 10 3 = [ OH] and the assumption is valid (1.4%) poh = log 10 [ OH] = 2.26 ph = 14 poh = 11.74 0.400-x (c) We can use the Henderson-Hasselbalch equation to solve this question: ph = pka + log 10 [conjugate base] [acid]
All we need to do is determine the concentrations of N(CH 3 ) 3 H + and N(CH 3 ) 3 : The total solution volume after mixing will be 0.200 L + 0.150 L = 0.350 L [N(CH 3 ) 3 ] = initial moles = (0.200 L)(0.300 mol/l) = 0.060 mol = 0.171 M total volume 0.350 L 0.350 L [N(CH 3 ) 3 H + ] = initial moles = (0.150 L)(0.300 mol/l) = 0.045 mol = 0.129 M total volume 0.350 L 0.350 L ph = 9.88 + log 10 (0.171) = 10.0 (0.129) (Note: you can also solve this problem with the use of an ice table format, and you will get the same answer). (d) Since both N(CH 3 ) 3 and N(CH 3 ) 3 H + are diluted equally when the dilution to 2.0 L takes place, their mole ratio does not change, and it is that ratio that causes changes in ph, according to the Henderson-Hasselbalch equation. Since the moles of N(CH 3 ) 3 and N(CH 3 ) 3 H + are unchanged, the concentration of [ OH] and [H 3 O + ], and thus ph, should remain unchanged. 7. (a) Calculate the ph of the aqueous solution obtained on dissolving 0.22 mole sodium hypochlorite (Na + ClO ) in 1.00 L of water. K a (HClO) = 3.5 x 10 8. (b) Calculate the ph of a 1.00 L solution containing 0.0500 mole hypochlorous acid and 0.200 mole sodium hypochlorite. (c) To the solution in part (b) is added 3.65 g of HCl gas. What is the ph after the addition? (molecular mass of HCl = 36.5 g/mol). Assume no volume change on addition of HCl(g). (a) ClO + H 2 O Ξ HOCl + OH Initial 0.22 M 0 0 (+10 7 M) Change -x +x +x Equilibrium 0.22-x x x K b = K w = 1.0 10 14 = 2.9 10 7 = [HOCl][ OH]
K a 3.5 10 8 [ClO ] K b = (x)(x) (0.22-x) Assume x is small, such that x << 0.22, then (0.22-x) 0.22 K b = x 2 = 2.9 10 7 0.22 x = 2.53 10 4 M = [ OH] Check that the assumption is valid, [(2.53 10 4 )/(0.22)] 100% = 0.1 % assumption okay poh = log 10 [ OH] = 3.60 ph = 14 poh = 10.4 (b) ClO + H 2 O Ξ HOCl + OH Initial 0.20 M 0.050 0 (10 7 M from water) Change -x +x +x Equilibrium 0.20-x x + 0.050 x We have the same Kb expression in this case, so it will look like: K b = (x)(0.050 + x) (0.20-x) Again make the assumption that x << 0.050 and x << 0.20, then we can write K b = (x)(0.050) = 2.9 10 7 (0.20) x = 1.16 10 6 = [OH ] Check that the assumption is valid, [(1.16 10 6 )/(0.20)] 100% = <1 %. Assumption okay.
poh = log 10 [OH ] = 5.94 ph = 14 poh = 8.06 Alternatively, one can use the Henderson-Hasselbalch equation to solve this question: ph = pka + log 10 [conjugate base] [acid] ph = 7.46 + log 10 (0.20) = 8.06 (0.050) (c) Addition of HCl: amount added = 3.65 g/(36.5 g/mol) = 0.10 mol HCl The added HCl consumes 0.10 mol ClO and produces 0.10 mol HOCl. H 3 O + + ClO H 2 O + HOCl We initially had 0.200 M ClO (from 0.200 mol in 1.00 L). If we use up 0.100 mol, then 0.100 mol remains, and so the new concentration is: [ClO ] = 0.100 M Likewise, we initially had 0.0500 M HOCl. If we make 0.100 mol, then the new concentration is: [HOCl] = 0.150 We now have a new equilibrium between HOCl and ClO. ClO + H 2 O Ξ HOCl + OH We can determine the new ph of the system by using the "ICE" table approach, or by using the Henderson-Hasselbalch equation: poh = pk b + log 10 [conjugate acid] [base] poh = 6.54 + log 10 (0.150) (0.100) poh = 6.72
ph = 14-6.72 = 7.28 As we expected, the ph of the buffer solution decreases only slightly after the addition of strong acid (HCl).