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