Lecture 18 Notes
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Transcript Lecture 18 Notes
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