Download presentation

Presentation is loading. Please wait.

Published byEleanore O’Neal’ Modified about 1 year ago

1
Chapter 8 Activity

2
Homework Chapter 8 - Activity 8.2, 8.3, 8.6, 8.9, 8.10, 8.12

3
Question 8.2 Q: Which statements are true? In the ionic strength, , 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!!

4
Question 8.3 Calculate the ionic strength of a) 0.0087 M KOH b) 0.0002 M La(IO 3 ) 3 (assuming complete dissociation at low concentration) Remember for +1/-1 systems: Ionic strength, = Molarity, M Ionic strength ( ) = ½ (c 1 z 1 2 + c 2 z 2 2 + …) = ½ [0.0087M(+1) 2 + 0.0087M(-1) 2 ] = 0.0087 M

5
Question 8.3 (cont’d) Calculate the ionic strength of a) 0.0087 M KOH b) 0.0002 M La(IO 3 ) 3 (assuming complete dissociation at low concentration) Ionic strength ( ) = ½ (c 1 z 1 2 + c 2 z 2 2 + …) = ½ [0.0002M(+3) 2 + 0.006M(-1) 2 ] = 0.001 2 M

6
Question 8.6 Calculate the activity coefficient of Zn 2+ when M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1. =-0.375 =0.42 2 a)

7
Question 8.6 (cont’d) Calculate the activity coefficient of Zn 2+ when M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1. = 0.43 2

8
Question 8-9 0.00100 M KNO 3. Calculate the concentration of Hg 2 2+ in saturated solutions of Hg 2 Br 2 in 0.00100 M KNO 3. Hg 2 Br 2(s) Hg 2 2+ + 2Br - K sp =5.6x10 -23 ICE some- - -x+x+2x some-x+x+2x

9
8-10. Find the concentration of Ba 2+ in a 0.100 M (CH 3 ) 4 NIO 3 solution saturated with Ba(IO 3 ) 2 (s).+ Ba(IO 3 ) 2 Ba 2+ + 2IO 3 K sp = 7.11 x 10 -11 ICE some- 0.100 -x+x+2x some-x+x+2x

10

11

12
8-10. Find the concentration of Ba 2+ in a 0.100 M (CH 3 ) 4 NIO 3 solution saturated with Ba(IO 3 ) 2 (s).+ Ba(IO 3 ) 2 Ba 2+ + 2IO 3 K sp = 1.5 x 10 -9 ICE some- 0.1 -x+x+2x some-x+x0.1+2x

13
8-10. Find the concentration of Ba 2+ in a 0.100 M (CH 3 ) 4 NIO 3 solution saturated with Ba(IO 3 ) 2 (s).+ Ba(IO 3 ) 2 Ba 2+ + 2IO 3 K sp = 1.5 x 10 -9 ICE some- 0.1 -x+x+2x some-x+x0.1+2x X = 6.5 7 x 10 -7

14
Question 8-12 correctly, Using activities correctly, calculate the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO 3. What is the pH if you neglected activities? ( ) = ½ (c 1 z 1 2 + c 2 z 2 2 + …) = ½ [0.010M(+1) 2 + 0.010M(-1) 2 + 0.0120M(+1) 2 + 0.0120M(-1) 2 ] = 0.022 0 M OH = 0.873pH = A H = [H + ] H

15
Question 8-12 (cont’d) correctly, Using activities correctly, calculate the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO 3. What is the pH if you neglected activities? pH = 11.94

16
Question 8-12 (cont’d) correctly, Using activities correctly, calculate the pH of a solution containing 0.010 M NaOH plus 0.0120 M LiNO 3. What is the pH if you neglected activities? pH ~ -log[H + ] =

17
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.

18
Acid/Base Titrations

19
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

20
Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr First -find Volume at equivalence M 1 V 1 = M 2 V 2 (0.050 L)(0.02000M) = 0.1000 V V = 10.0 mL

21
Strong Acid/Strong Base 50.00 mL of 0.02000 M KOH Titrated with 0.1000 M HBr Second – find initial pH pH = - logA H ~ -log [H + ] pOH = -logA OH ~ -log [OH - ] pH = 12.30

22
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) -> H 2 O (l) + KBr(aq) Before After 0.001000 mol0.0006000 mol 0.000400 mol0 mol Limiting Reactant 0.0006000 mol pH = 11.8 (~6 ml)

23
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) -> H 2 O (l) + KBr(aq) Before After 0.001000 mol0.0010000 mol 0 mol 0.0010000 mol pH = 7.0

24
Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr Finally – find pH after equivalence point KOH (aq) + HBr (aq) -> H 2 O (l) + KBr(aq) Before After 0.001000 mol0.001200 mol 0 mol0.0002000 mol 0.0010000 mol pH = 2.5 12 ml Limiting Reactant

25

26

27
Titration of WEAK acid with a strong base

28

29
Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid K a = 1.8 x 10 -5 Titrant = 0.100 M NaOH First, calculate the volume at the equivalence-point M 1 V 1 = M 2 V 2 (0.0250 L) 0.1000 M = 0.1000 M (V 2 ) V 2 = 0.0250 L or 25.0 mL

30
Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid K a = 1.8 x 10 -5 Titrant = 0.100 M NaOH Second, Calculate the initial pH of the acetic acid solution

31
Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid K a = 1.8 x 10 -5 Titrant = 0.100 M NaOH Third, Calculate the pH at some intermediate volume

32
Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid K a = 1.8 x 10 -5 Titrant = 0.100 M NaOH Fourth, Calculate the pH at equivalence

33
Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid K a = 1.8 x 10 -5 Titrant = 0.100 M NaOH Finally calculate the pH after the addition 26.0 mL of NaOH

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google