Download presentation

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, m, range of 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**

M KOH M La(IO3)3 (assuming complete dissociation at low concentration) Ionic strength (m) = ½ (c1z12+ c2z22 + …) = ½ [0.0087M(+1) M(-1)2] = M Remember for +1/-1 systems: Ionic strength, m = Molarity, M

5
**Question 8.3 (cont’d) Calculate the ionic strength of**

M KOH M La(IO3)3 (assuming complete dissociation at low concentration) Ionic strength (m) = ½ (c1z12+ c2z22 + …) = ½ [0.0002M(+3) M(-1)2] = M

6
Question 8.6 Calculate the activity coefficient of Zn2+ when m = M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1. a) =-0.375 g=0.422

7
Question 8.6 (cont’d) Calculate the activity coefficient of Zn2+ when m = M by using (a) Equation 8-6 and (b) linear interpolation with Table 8-1. = 0.432

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

9
8-10. Find the concentration of Ba2+ in a 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 -x +x +2x some-x +x +2x

12
8-10. Find the concentration of Ba2+ in a 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 -x +x +2x some-x +x 0.1+2x

13
8-10. Find the concentration of Ba2+ in a 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 -x +x +2x some-x +x 0.1+2x X = 6.57 x 10-7

14
Question 8-12 Using activities correctly, calculate the pH of a solution containing M NaOH plus M LiNO3. What is the pH if you neglected activities? (m) = ½ (c1z12+ c2z22 + …) = ½ [0.010M(+1) M(-1) M(+1) M(-1)2] = M gOH = pH = AH = [H+]gH

15
Question 8-12 (cont’d) Using activities correctly, calculate the pH of a solution containing M NaOH plus M LiNO3. What is the pH if you neglected activities? pH = 11.94

16
Question 8-12 (cont’d) Using activities correctly, calculate the pH of a solution containing M NaOH plus M LiNO3. 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 M KOH Titrated with M HBr First -find Volume at equivalence M1V1 = M2V2 (0.050 L)( M) = V V = 10.0 mL

21
**Strong Acid/Strong Base**

50.00 mL of M KOH Titrated with M HBr Second – find initial pH pH = - logAH ~ -log [H+] pOH = -logAOH ~ -log [OH-] pH = 12.30

22
**Strong Acid/Strong Base**

50 mL of M KOH Titrated with M HBr Third– find pH at mid-way volume KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) (~6 ml) Before After mol mol mol 0 mol Limiting Reactant mol mol pH = 11.8

23
**Strong Acid/Strong Base**

50 mL of M KOH Titrated with M HBr Fourth – find pH at equivalence point KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) Before After mol mol 0 mol 0 mol mol mol pH = 7.0

24
**Strong Acid/Strong Base**

50 mL of M KOH Titrated with M HBr Finally – find pH after equivalence point 12 ml KOH (aq) + HBr (aq) -> H2O (l) + KBr(aq) Limiting Reactant Before After mol mol 0 mol mol mol pH = 2.5

27
**Titration of WEAK acid with a strong base**

29
**Titration of a weak acid solution with a strong base.**

25.0 mL of M acetic acid Ka = 1.8 x 10-5 Titrant = M NaOH First, calculate the volume at the equivalence-point M1V1 = M2V2 ( L) M = M (V2) V2 = L or 25.0 mL

30
**Titration of a weak acid solution with a strong base.**

25.0 mL of M acetic acid Ka = 1.8 x 10-5 Titrant = 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 M acetic acid Ka = 1.8 x 10-5 Titrant = 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 M acetic acid Ka = 1.8 x 10-5 Titrant = M NaOH Fourth, Calculate the pH at equivalence

33
**Titration of a weak acid solution with a strong base.**

25.0 mL of M acetic acid Ka = 1.8 x 10-5 Titrant = M NaOH Finally calculate the pH after the addition mL of NaOH

Similar presentations

Presentation is loading. Please wait....

OK

Yesterday’s Homework Page 611 # 19 Page 612 # 20.

Yesterday’s Homework Page 611 # 19 Page 612 # 20.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on poultry farm management Ppt on applied operational research techniques Ppt on standing order definition Ppt on energy giving food proteins Ppt on asian continent outline Ppt on event handling in javascript 17%5 Ppt on web page designing Free ppt on mobile number portability ppt Ppt on trade union act 1926 in india Ppt on principles of peace building activities