CHEM 213 Chemical Analysis Exam 2 Tuesday May 11, 2004 1 10_ (of 10) 2 10_ (of 10) 3 10_ (of 10) 4 10_ (of 10) 5 10_ (of 10) 6 10_ (of 10) 7 20_ (of 20) 8 10_ (of 10) 9 10_ (of 10) Σ 100 100% KEY Name: (please print)
1. Calculate the solubility of PbI 2 (K sp = 3.0 10-16 ) in a 0.0333 M solution of Mg(ClO 4 ) 2 using a. molar concentrations. (5 points) K sp = [Pb 2+ ][I - ] 2 = 3.0 10-16 [Pb 2+ ] = s, [I - ] = 2s K sp = s(2s) 2 = 3.0 10-16 s = solubility = (3.0 10-16 /4) 1/3 = 4.22 10-6 b. activities. (you find Kjelland s table of activity coefficients on page 11 of this exam). (5 points) µ = ½ Σc i Z i 2 = ½ [0.0333(2) 2 + 0.0666(1) 2 ] = 0.100 γ Pb2+ = 0.36, γ I- = 0.75 K sp = [Pb 2+ ][I - ] 2 γ Pb2+ ( γ I- ) 2 K sp = K sp /(γ Pb2+ ( γ I- ) 2 ) = 3.0 10-16 /(0.36 0.75 2 ) = 1.48 10-15 [Pb 2+ ] = s, [I - ] = 2s K sp = s(2s) 2 = 1.48 10-15 s = solubility = (1.48 10-15 /4) 1/3 = 7.18 10-6 2
2. A 40.0-mL solution of 0.040 M Hg 2 (NO 3 ) 2 was titrated with 60.0 ml of 0.100 M KI to precipitate Hg 2 I 2 (K sp = 1.1 10-28 ). a. What volume of KI solution is needed to reach the equivalence point. (2 points) moles of I - = 2(moles of Hg 2 2+ ) (V c )(0.100 m) = 2(40.0 ml) 0.040 M) V c = 32.0 ml b. Calculate the ionic strength (µ) of the solution when 60.0 ml of KI have been added. (4 points) Vitually all the Hg 2 2+ has precipitated, along with 3.20 mmol of I -. The ions remaining in solution are: [NO 3 - ] = 3.20 mmol/100 ml = 0.0320 M [I - ] = 2.80 mmol/100 ml = 0.0280 M [K + ] = 6.00 mmol/100 ml = 0.0600 M µ = ½ Σc i Z i 2 = ½ [0.032(1) 2 + 0.0280(1) 2 + 0.0600(1) 2 ] = 0.060 M c. Using activities (activity coefficients on page 11), calculate phg 2 2+ (= -log a Hg2 2+). (4 points) a Hg2 2+ = K sp /(a I- ) 2 = 1.1 10-28 /((0.81) 2 (0.028) 2 ) = 2.14 10-25 phg 2 2+ = 24.67 3
3. (a) Find the ph of a solution prepared by dissolving 12.43 g of tris (MW = 121.135 g/mol) plus 4.67 g of tris hydrochloride (MW = 157.596 g/mol, K a = 8.4 10-9 ) in 1.00 L of water. (5 points) [B] = 12.43 g L -1 /121.135 g mol -1 = 0.1026 M [BH + ] = 4.67 g L -1 /157.596 g mol -1 = 0.0296 M ph = pk a + log [B]/[BH + ] = 8.075 + log(0.1026/0.0296) = 8.61 (b) If we add 12.0 ml of 1.00 M HCl to the solution made in (a), what will be the new ph? (5 points) When a strong acid is added to a weak base, both react completely to give BH +. We are adding 12.0 ml of 1.00 M HCl, which contains (0.0120 L)(1.00 mol/l) = 0.0120 mol of H +. This much H + will consume 0.0120 mol of B to create 0.0120 mol of BH +, which is shown in the table below: B + H + BH + Initial moles 0.1026 0.0120 0.0296 Final moles 0.0906-0.0416 ph = pk a + log[b]/[bh + ] = 8.075 + log(0.0906/0.0416) = 8.41 NOTE: The volume of the solution is in this case irrelevant (calculate the numbers with M instead of mol, and you will see), because volume cancels in the numerator and denominator of the log term: ph = pk a + log[(moles of B/L of solution)]/[(moles of BH + /L of solution)] 4
4. Calculate the concentration of CdS (K sp = 1 10-27 ) in a solution in which the [H 3 O + ] concentration is held constant at 1.0 10-4 M. Other equilibria that need to be considered to solve this problem are: H 2 S(aq) + H 2 O(l) HS - (aq) + H 3 O + (aq) K 1 = 9.6 10-8 HS - (aq) + H 2 O(l) S 2- (aq) + H 3 O + (aq) K 2 = 1.3 10-14 H 2 S(aq) + 2H 2 O(l) S 2- (aq) + 2H 3 O + K 1 K 2 = 1.25 10-21 For full credit, I require more than a formula copied from your cheat sheet. I need to see how you obtained this formula. (10 points) S = solubility = [Cd 2+ ] = [S 2- ] + [HS - ] + [H 2 S] K sp = [Cd 2+ ][S 2- ] (1) K 2 = [H 3 O + ][S 2- ]/[HS - ] = 1.3 10-14 (2) K 1 K 2 = [H 3 O + ] 2 [S 2- ]/[H 2 S] = 1.25 10-21 (3) From mass-balance consideration [Cd 2+ ] = [S 2- ] + [HS - ] + [H 2 S] (4) Substituting equations (2) and (3) into (4) gives [Cd 2+ ] = [S 2- ] + [H 3 O + ][S 2- ]/K 2 + [H 3 O + ] 2 [S 2- ]/K 1 K 2 = [S 2- ](1 + [H 3 O + ]/K 2 + [H 3 O + ] 2 /K 1 K 2 ) [S 2- ] = [Cd 2+ ]/(1 + [H 3 O + ]/K 2 + [H 3 O + ] 2 /K 1 K 2 ) Substituting [S 2- ] in equation (1) yields [Cd 2+ ] = (K sp /[Cd 2+ ])(1 + [H 3 O + ]/K 2 + [H 3 O + ] 2 /K 1 K 2 ) [Cd 2+ ] = (K sp (1 + [H 3 O + ]/K 2 + [H 3 O + ] 2 /K 1 K 2 )) 1/2 Inserting the actual values for K sp, K 2, K 1 K 2, and [H 3 O + ] [Cd 2+ ] = (1 10-27 (1 + 1.0 10-4 /1.3 10-14 + (1.0 10-4 ) 2 /1.25 10-21 )) 1/2 [Cd 2+ ] = 8.9 10-8 M 5
5. What is the ph of a solution that is prepared by dissolving 3.00 g of salicylic acid, C 6 H 4 (OH)COOH (138.12 g/mol, K a = 1.06 10-3 ) in 50.0 ml of 0.1130 M NaOH and diluting to 500.0 ml? Hint: The simplified Henderson-Hasselbach equation will not work in this case. (10 points) Original amount HA = 3.00 g (mmol HA/0.138.12 g) = 21.72 mmol Original amount NaOH = 50.0 ml (0.1130 mol/ml) = 5.65 mmol c HA = (21.72 5.65)mmol/500 ml= 3.214 10-2 M c NaA = 5.65 mmol/500 ml = 1.130 10-2 M [H 3 O + ] = K a [HA]/[A - ] = 1.06 10-3 x 3.214 10-2 /1.130 10-2 = 3.015 10-3 ph = 2.521 However, [H 3 O + ] is not << c HA and c NaA [HA] = 3.214 10-2 [H 3 O + ] + [OH - ] [A - ] = 1.130 10-2 + [H 3 O + ] - [OH - ] [OH - ] at this ph will be negligible. 1.06 10-3 = [H 3 O + ](1.130 10-2 + [H 3 O + ]0/(3.214 10-2 [H 3 O + ]) Rearranging gives [H 3 O + ] 2 + 1.236 10-2 [H 3 O + ] 3.07 10-5 = 0 [H 3 O + ] = 2.321 10-3 ph = 2.63 6
6. Calculate the equilibrium concentration of undissociated HCOOH in a formic acid solution with an analytical formic acid concentration of 0.0850 and a ph of 3.200 (K a (HCOOH) = 1.80 10-4 ). (10 points) α 0 = [H 3 O + ]/([H 3 O + ] + K a ) = 6.310 10-4 /(6.310 10-4 + 1.80 10-4 ) = 0.778 [HCOOH]/C T = [HCOOH]/0.0850 = α 0 [HCOOH] = 0.778 0.0850 = 6.61 10-2 M 7
7. Consider the titration of 50.0 ml of 0.050 M malonic acid (K a1 = 1.42 10-3, K a2 = 2.01 10-6 ) with 0.100 M NaOH. The titration reaction occurring is: HO 2 CCH 2 CO 2 H + OH - - O 2 CCH 2 CO 2 H + H 2 O - O 2 CCH 2 CO 2 H + OH - - O 2 CCH 2 CO 2 - + H 2 O Designate malonic acid as H 2 M and determine the ph after the following volumina of NaOH have been added: a. At 0.0 ml (3 points) H 2 M H + + HM - 0.050 x x x x 2 /(0.050 x) = K 1 x = 7.75 10-3 ph = 2.11 b. At 8.0 ml (3 points) H 2 M + OH - HM - + H 2 O Initial: 25 8 - - Final: 17 o 8 - ph = pk a1 + log([hm - ]/[H 2 M]) = 2.487 + log(8/17) = 2.52 c. At 12.5 ml (4 points) V b = ½ V e ph = pk a1 = 2.85 d. At 25.0 ml (4 points) At the first equivalence point, H 2 M has been converted to HM -. [H 3 O + ] = K a1 K a2 F + K a1 K w K a1 + F where F = (50/75)(0.050) = 0.0333 M [H 3 O + ] = 5.23 10-5 M ph = 4.28 e. At 50.0 ml (3 points) At the second equivalence point, H 2 M has been converted to M 2- : M 2- + H 2 O HM - + OH - (50/100)(0.050) x x x x 2 /(0.025 x) = K b1 = K w /K a2 x = 1.12 10-5 M ph = -log(k w /x) = 9.05 f. At 56.3 ml (3 points) There are 6.3 ml of excess NaOH [OH - ] = (6.3/106.3)(0.100) = 5.93 10-3 M ph = 11.77 8
8. Consider the titration of 50.00 ml of 0.1000 M HNO 2 (K a = 7.1 10-4 ) with 0.1000 M NaOH solution. What is the ph after the addition of: a. 0.00 ml (1 point) c HA = 0.1000 M [H 3 O + ] = [A - ] [HA] = 0.1000 [H 3 O + ] K a = 7.1 10-4 = [H 3 O + ][A - ]/[HA] = [H 3 O + ] 2 /(0.1000 - [H 3 O + ]) [H 3 O + ] 2 + 7.1 10-4 [H 3 O + ] 0.1000 7.1 10-4 = 0 ph = 2.09 b. 15.00 ml (2 points) c NaA = (15.00 0.1000)/50.00 +15.00) = 2.3 10-2 c HA = (50.00 0.1000) 15.00 0.1000)/65.00 = 5.3 10-2 This is a buffer, but simplifying assumptions are not valid. K a = [H 3 O + ](c NaA + [H 3 O + ] [OH - ])/( c HA - [H 3 O + ] + [OH - ]), rearranged: [H 3 O + ] 2 + (K a + c NaA ) [H 3 O + ] - K a c HA = 0 [H 3 O + ] 2 + 0.02371[H 3 O + ] 3.763 10-5 = 0 ph = 2.83 c. 25.00 ml (2 points) c NaA = (25.00 0.1000)/50.00 +25.00) = 3.33 10-2 c HA = (50.00 0.1000) 25.00 0.1000)/75.00 = 3.33 10-2 This is still a buffer, use above equation. [H 3 O + ] 2 + (K a + c NaA ) [H 3 O + ] - K a c HA = 0 [H 3 O + ] 2 + 0.0340433[H 3 O + ] 2.367 10-5 = 0 ph = 3.17 d. 35.00 ml ( 2 points) c NaA = (35.00 0.1000)/50.00 +35.00) = 4.12 10-2 c HA = (50.00 0.1000) 35.00 0.1000)/85.00 = 1.76 10-2 This is still a buffer, use above equation. [H 3 O + ] 2 + (K a + c NaA ) [H 3 O + ] - K a c HA = 0 [H 3 O + ] 2 + 0.04191[H 3 O + ] 1.25 10-5 = 0 ph = 3.53 e. 50.00 ml (2 points) This is the equivalence point and we have made a solution of NaNO 2. c NaNO2 = 50.00 0.100/100 ml = 0.05000 M K b = K w /K a = 10-14 /7.1 10-4 = 1.408 10-11 [OH - ] = (K b c NaNO2 ) 1/2 = 8.390 10-7 poh = 6.08 ph = 7.92 f. 51.00 ml (1 point) We now have an excess of NaOH. [OH - ] c NaOH = 51.00 0.1000 50.00 0.1000)/101.0 = 9.901 10-4 poh = 3.00 ph = 11.00 9
9. How would you prepare 1.00 L of a buffer with ph of 6.00 from 0.500 Na 3 AsO 4 and 0.400 M HCl? K a1 (H 3 AsO 4 ) = 5.8 10-3, K a2 (H 3 AsO 4 ) = 1.1 10-7, K a3 (H 3 AsO 4 ) = 3.2 10-12 (10 points) ph = 6.00 [H 3 O + ] = antilog (-6.00) = 1.00 10-6 [H 3 O + ][HAsO 4 2- ]/[H 2 AsO 4 - ] = 1.1 10-7 [HAsO 4 2- ]/[H 2 AsO 4 - ] = 1.1 10-7 /1.00 10-6 = 0.11 (1) Let V H3AsO4 and V NaOH be the volume in milliliters of the two reagents. Then: V H3AsO4 + V NaOH = 1000 ml (2) From mass-balance consideration, we may write that in the 1000 ml, no. mmol NaH 2 AsO 4 + no. mmol Na 2 HAsO 4 = 0.500 V Na3AsO4 (3) 2 no. mmol NaH 2 AsO 4 + no. mmol Na 2 HAsO 4 = 0.400 V HCl (4) Equation (1) can be written as: no. mmol Na 2 HAsO 4 /1000 no. mmol NaH 2 AsO 4 /1000 no. mmol Na 2 HAsO = 4 = 0.11 (5) no. mmol NaH 2 AsO 4 Thus we have four equations: (2), (3), (4), and (5) and four unknowns: V Na3AsO4, V HCl, no. mmol NaH 2 AsO 4 and no. mmol of Na 2 HAsO 4. Subtracting equation (3) from (4) yields: no. mmol NaH 2 AsO 4 = 0.400 V HCl 0.500 V Na3AsO4 (6) Substituting equation (6) into (3) gives no. mmol Na 2 HPO 4 + 0.400 V HCl 0.500 V Na3AsO4 = 0.500 V Na3AsO4 no. mmol Na 2 HPO 4 = 0.400 V HCl + 1.000 V Na3AsO4 (7) Substituting equations (6) and (7) into (5) gives ( 0.400 V HCl + 1.000 V Na3AsO4 )/(0.400 V HCl 0.500 V Na3AsO4 ) = 0.11 This equation rearranges to: 0.444 V HCl = 1.055 V Na3AsO4 Substituting equation (2) gives 0.444(1000 - V Na3AsO4 ) = 1.055 V Na3AsO4 V Na3AsO4 = 444/1.499 = 296 ml V HCl = 1000 296 = 704 ml 10
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