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 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 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. a) =-0.375 g=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. = 0.432

Question 8-9 Calculate the concentration of Hg22+ in saturated solutions of Hg2Br2 in 0.00100 M KNO3. Hg2Br2(s) D Hg22+ + 2Br- Ksp=5.6x10-23 some - - -x +x +2x some-x +x +2x I C E

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 Ksp = 7.11 x 10-11 I C E some - 0.100 -x +x +2x some-x +x +2x

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 Ksp = 1.5 x 10-9 I C E some - 0.1 -x +x +2x some-x +x 0.1+2x

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 Ksp = 1.5 x 10-9 I C E some - 0.1 -x +x +2x some-x +x 0.1+2x X = 6.57 x 10-7

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 gOH = 0.873 pH = AH = [H+]gH

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 = 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+] =

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 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr Third– find pH at mid-way volume KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) (~6 ml) Before After 0.001000 mol 0.0006000 mol 0.000400 mol 0 mol Limiting Reactant 0.0006000 mol 0.0006000 mol 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 KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) Before After 0.001000 mol 0.0010000 mol 0 mol 0 mol 0.0010000 mol 0.0010000 mol pH = 7.0

Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr Finally – find pH after equivalence point 12 ml KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) Limiting Reactant Before After 0.001000 mol 0.001200 mol 0 mol 0.0002000 mol 0.0010000 mol pH = 2.5

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