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Chapter 8 Activity Homework Chapter 8 - Activity 8.2, 8.3, 8.6, 8.9, 8.10, 8.12 Question 8.2 Q: Which statements are true? In the ionic strength, m, range of 0-0.1 M, activity coefficients decrease with: a) increasing ionic strength b) increasing ionic charge c) decreasing hydrated radius All are true!! Question 8.3 Calculate the ionic strength of a) b) 0.0087 M KOH 0.0002 M La(IO3)3 (assuming complete dissociation at low concentration) Ionic strength (m) = ½ (c1z12+ c2z22 + …) = ½ [0.0087M(+1)2+ 0.0087M(-1)2] = 0.0087 M Remember for +1/-1 systems: Ionic strength, m = Molarity, M Question 8.3 (cont’d) Calculate the ionic strength of a) b) 0.0087 M KOH 0.0002 M La(IO3)3 (assuming complete dissociation at low concentration) Ionic strength (m) = ½ (c1z12+ c2z22 + …) = ½ [0.0002M(+3)2+ 0.006M(-1)2] = 0.0012 M Question 8.6 Calculate the activity coefficient of Zn2+ when m = 0.083 M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1. 0.51z 2x m log x = a) 1 3.3 x m log x = 0.51(2) 2 0.083 (600) 0.083 1 305 =-0.375 =0.422 Question 8.6 (cont’d) Calculate the activity coefficient of Zn2+ when m = 0.083 M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1. unknowny interval known x interval = y x 0.083 0.05 = (0.18 0.245 ) 0.245 0.1 0.05 = 0.432 Question 8-9 Calculate the concentration of Hg22+ in saturated solutions of Hg2Br2 in 0.00100 M KNO3. Hg2Br2(s) D Hg22+ + 2BrI C E some -x some-x +x +x +2x +2x Ksp=5.6x10-23 2 Ksp = AHg ABr = 5.6 1023 2 Ksp = [Hg22 ] Hg [Br ]2Br Br = 5.6 1023 23 Ksp = [ x]0.867[2x] 0.964 = 5.6 10 [ x] = [ Hg22 ] = 2.6 108 M 2 2 8-10. Find the concentration of Ba2+ in a 0.100 M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+ Ba(IO3)2 Ba2+ + 2IO3 I C E some -x some-x +x +x K sp = ABa2 A 2 IO3 Ksp = 7.11 x 10-11 0.100 +2x +2x = [ Ba ] Ba2 [ IO ] 2 2 3 2 IO3 8-10. Find the concentration of Ba2+ in a 0.100 M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+ Ba(IO3)2 Ba2+ + 2IO3 I C E some -x some-x +x +x K sp =A ABa22 A A K sp = Ba Ksp = 1.5 x 10-9 0.1 +2x 0.1+2x 2 = [ Ba ] [ IO ] = [ Ba ]0.Ba 382[ IO ] 0.775 22 IO3 IO 3 2 IO3 22 22 33 2 IO3 Ksp = [x ]02.38 ABa A [0.=1][ x0].0775 .38[2 x 0.1] 0.775 2 2 2 2 8-10. Find the concentration of Ba2+ in a 0.100 M (CH3)4NIO3 solution saturated with Ba(IO3)2 (s).+ Ba(IO3)2 Ba2+ + 2IO3 I C E some -x some-x +x +x Ksp = 1.5 x 10-9 0.1 +2x 0.1+2x Ksp = [x]0.38[0.1] 0.775 2 X = 6.57 x 10-7 2 Question 8-12 Using activities correctly, calculate the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO3. What is the pH if you neglected activities? (m) = ½ (c1z12+ c2z22 + …) = ½ [0.010M(+1)2+ 0.010M(-1)2+ 0.0120M(+1)2+ 0.0120M(-1)2] = 0.0220 M OH = 0.873 pH = AH = [H+]H Question 8-12 (cont’d) Using activities correctly, calculate the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO3. What is the pH if you neglected activities? Kw = AH AOH = 1.0x1014 14 Kw 1.0 x10 AH = = AOH [OH ] OH Kw 1.0 x1014 AH = = AOH [0.010](0.873) AH = 1.151012 pH = 11.94 Question 8-12 (cont’d) Using activities correctly, calculate the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO3. What is the pH if you neglected activities? pH ~ -log[H+] Kw = 12.00 = log [OH ] Finally Calculate the pH of a solution that contains 0.1 M Acid and 0.01 M conjugate base Calculate the pH of a solution containing 0.1 M Acid and 0.05 M conjugate base. Calculate the pH of a solution containing 0.1 M Acid and 0.1 M conjugate base. Acid/Base Titrations Titrations Titration Curve – always calculate equivalent point first Strong Acid/Strong Base Regions that require “different” calculations B/F any base is added Half-way point region At the equivalence point After the equivalence point Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr First -find Volume at equivalence M1V1 = M2V2 (0.050 L)(0.02000M) = 0.1000 V V = 10.0 mL Strong Acid/Strong Base 50.00 mL of 0.02000 M KOH Titrated with 0.1000 M HBr Second – find initial pH pH = - logAH ~ -log [H+] pOH = -logAOH ~ -log [OH-] pH = 12.30 Strong Acid/Strong Base Before After 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr (~6 ml) Third– find pH at mid-way volume KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) 0.001000 mol 0.000400 mol 0.0006000 mol 0 mol 0.0006000 mol 0.0006000 mol Limiting Reactant pH = 11.8 Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr Fourth – find pH at equivalence point Before After KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) 0.001000 mol 0 mol pH = 7.0 0.0010000 mol 0 mol 0.0010000 mol 0.0010000 mol Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr Finally – find pH after equivalence point Before After 12 ml KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) 0.001000 mol 0.001200 mol 0 mol 0.0002000 mol pH = 2.5 Limiting Reactant 0.0010000 mol Typical pH titration 14 12 pH 10 8 6 4 2 0 0 5 10 mL of HBr 15 20 Titration of WEAK acid with a strong base Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH First, calculate the volume at the equivalence-point M1V1 = M2V2 (0.0250 L) 0.1000 M = 0.1000 M (V2) V2 = 0.0250 L or 25.0 mL Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH Second, Calculate the initial pH of the acetic acid solution Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH Third, Calculate the pH at some intermediate volume Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH Fourth, Calculate the pH at equivalence Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid Ka = 1.8 x 10-5 Titrant = 0.100 M NaOH Finally calculate the pH after the addition 26.0 mL of NaOH