CHEMISTRY 1AA3 Tutorial 2 Answers - WEEK E WEEK OF JANUARY 22, (i) What is the conjugate base of each of the following species?

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1 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 What was the concentration of Ba(OH) 2 in the original 20 ml solution? (a) Ba(OH) 2 Ba OH M 0.04M 0.08M [OH - ] = 0.08M and poh = 1.10

2 ph = 14 - poh = = (b) ph = poh = = 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 (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: 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

3 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 ] = 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 EQUATION: (CH 3 ) 2 NH + H 2 O Ξ (CH 3 ) 2 NH 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: 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.]

4 0.50 poh = -log[oh - ] = -log(1.92 x 10-2 ) = 1.72 and ph = poh = = Percent dissociation = 1.92 x 10-2 x 100% = 3.8% 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 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 g/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 ] = [H 3 O + ] = M K a = (9.77 x 10-4 )( 9.77 x 10-4 ) = 1.95 x (a) A M solution of trimethylammonium chloride, N(CH 3 ) 3. HCl, has a ph of Calculate the values of K b for trimethylamine, N(CH 3 ) 3. (b) What is the ph of M (CH 3 ) 3 in water? (c) What is the ph of the solution resulting from the mixing of 200 ml of M N(CH 3 ) 3 and 150 ml of 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 +

5 Initial M 0 0 Change -x +x +x Equilibrium x x x In this case we are given information about the equilibrium concentrations of these species, since we are told that ph = This means that at equilibrium, [H 3 O + ] = antilog (ph) = = M This is the value for x in our table, and so we can write K a = [N(CH 3 ) 3 ][ H 3 O + ] = ( ) 2 = [N(CH3)3H + ] ( ) K b = K w = = K a (b) This is just a weak base problem: N(CH 3 ) 3 + H 2 O Ξ N(CH 3 ) 3 H + + OH Initial M 0 0 Change -x +x +x Equilibrium x x x K b = = [N(CH 3 ) 3 H + ][ OH] = x 2 Assuming x << 0.400, this solves to give [N(CH3)3] x = = [ OH] and the assumption is valid (1.4%) poh = log 10 [ OH] = 2.26 ph = 14 poh = x (c) We can use the Henderson-Hasselbalch equation to solve this question: ph = pka + log 10 [conjugate base] [acid]

6 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 L L = L [N(CH 3 ) 3 ] = initial moles = (0.200 L)(0.300 mol/l) = mol = M total volume L L [N(CH 3 ) 3 H + ] = initial moles = (0.150 L)(0.300 mol/l) = mol = M total volume L L ph = 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 (b) Calculate the ph of a 1.00 L solution containing mole hypochlorous acid and 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 = = = [HOCl][ OH]

7 K a [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 = x = M = [ OH] Check that the assumption is valid, [( )/(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 (10 7 M from water) Change -x +x +x Equilibrium 0.20-x x x We have the same Kb expression in this case, so it will look like: K b = (x)( x) (0.20-x) Again make the assumption that x << and x << 0.20, then we can write K b = (x)(0.050) = (0.20) x = = [OH ] Check that the assumption is valid, [( )/(0.20)] 100% = <1 %. Assumption okay.

8 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 = 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 M ClO (from mol in 1.00 L). If we use up mol, then mol remains, and so the new concentration is: [ClO ] = M Likewise, we initially had M HOCl. If we make mol, then the new concentration is: [HOCl] = 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 = log 10 (0.150) (0.100) poh = 6.72

9 ph = = 7.28 As we expected, the ph of the buffer solution decreases only slightly after the addition of strong acid (HCl).