Transcript Hein and Arena

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1
2
I. Introduction
A. A reaction at “equilibrium” is able to
react in both directions:
A + B <=> C + D
B. At equilibrium, the rates of the
forward and reverse reactions are
equal, and the concentrations of the
reactants and products are constant.
3
C. The equilibrium constant (K) is a value
representing the unchanging concentrations of
the reactants and the products in a chemical
reaction at equilibrium.
D. The equilibrium constant (K) is given a
subscript to indicate the type of reaction:
1) Ka = weak acid ionization
2) Kb = weak base ionization
3) Kc = concentrations of non acid/base
4) Kp = pressures of non acid/base
5) Ksp = “insoluble” solids that “dissolve”
6) Keq = generic “equilibrium” constant
4
II. Equilibrium Constants
A. For the general reaction:
aA + bB → cC + dD
at a given temperature ….
→
c
K eq
d
[C] [D]
=
a
b
[A] [B]
[ products ]
K
[reac tan ts]
ONLY (aq) and (g) ARE ALLOWED IN EXPRESSION !!!
5
Ex: Write the equilibrium expression for the
reaction:
3H + N → 2NH
→
2(g)
2 (g)
3 (g)
2
K eq
[NH3 ]
=
3
[H2 ] [N2 ]
7
Ex: Write the equlibrium expressionfor the
reaction:
4NH
+ 3O → 2N + 6H O
2 (g)
→
3 (g)
2(g)
2
K eq
2
(g)
6
[N2 ] [H2O]
=
4
3
[NH3 ] [O2 ]
8
Ex: Write the equlibrium expression for the
reaction:
2 KClO3 (s) <=> 2 KCl (s) + 3 O2 (g)
Keq = [O2
3
]
9
B. The magnitude of an equilibrium constant
indicates the extent to which the forward and
reverse reactions take place.
H2(g) + I2(g) → 2HI(g)
→
2
K eq
[HI]
o
=
= 54.8 at 425 C
[H2 ] [I2 ]
At
At equilibrium
equilibrium more
more product
reactant
than
exists.
than reactant
product exists.
COCl2(g) →CO(g) + Cl2(g)
→
K eq
[CO][Cl2 ]
-4
o
=
= 7.6 x 10 at 400 C
[COCl2 ]
10
C. When the molar concentrations of
species in an equilibrium reaction
known, the Keq can be calculated
substituting the concentrations into
equilibrium constant expression.
all
are
by
the
11
Calculate Keq for the following reaction if the
equilibrium concentrations of PCl5 = 0.030 mol/L,
PCl3 = 0.97 mol/L and Cl2 = 0.97 mol/L at 300oC.
PCl5(g ) → PCI3(g) + Cl2(g)
→
K eq
[PCl3 ][Cl2 ] (0.97)(0.97)
= 31
=
=
(0.030)
[PCl5 ]
12
Calculate Kp for the following reaction if the
equilibrium pressures of PCl5 = 1.46 atm, PCl3 = 2.50
atm and Cl2 = 2.50 atm at 300oC.
→

PCl5(g ) → PCI3(g) + Cl2(g)
( pPCl3 )(pCl2 )
Kp 
p(PCl5 )
Kp = 4.28
13
III. Calculating Equilibrium Concentrations
A. If K is known, then concentrations of all
species at equilibrium can be calculated
B.The procedure requires constructing an
ICE chart:
I = initial
C = change (pay attention to coefficients!)
E = equilibrium
C. Substitute the E values into the
equilibrium expression and solve for x
14
The equilibrium constant for the reaction below is 54.8
at 425 oC. If .500 moles each of H2(g) and I2(g) are
introduced into a 2.00 L container, what will be the
concentrations at equilibrium?
H2 (g) + I2 (g) <=> 2 HI (g)
I 0.250 0.250
-x
C -x
E (.250-x) (.250-x)
2
[HI]
Kc 
= 54.8
[H 2 ][I2 ]
+ 2x
(2x)
Solve for x and
substitute back in
15
The equilibrium constant for the reaction below is 54.8
at 425 oC. If .500 moles of H2(g) and 0.750 mole I2(g)
are introduced into a 1.50 L container, what will be the
concentrations of all species at equilibrium?
H2 (g) + I2 (g) <=> 2 HI (g)
I
C
E
16
A mixture of hydrogen and nitrogen in a reaction
vessel is allowed to attain equilibrium at 472 oC. The
equilibrium mixture of gases ws analyzed and found to
contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3.
From these data, calculate the Kp.?
3 H2 (g) + N2 (g) <=> 2 NH3 (g)
17
Enough ammonia is dissolved in 5.00 liters of water to
produce a solution that is 0.0124 M in ammonia. The
solution is then allowed to come to equilibrium.
Analysis of the equilibrium mixture shows that the
concentration of OH- is 4.64 x 10-6 M. Calculate Kc
for the reaction.
NH3 (aq) + H2O (l) <=> NH4+(aq)+ OH-(aq)
18
Sulfur trioxide decomposes at high temperatures in a
sealed container. Initially the vessel is filled with SO3
at a partial pressure of 0.500 atm. At equilibrium the
SO3 partial pressure is 0.200 atm. Calculate Kp.
2 SO3 (g) <=> 2 SO2 (g)+ O2(g)
19
Ionization Constants
20
In addition to Kw, several other ionization
constants are used.
21
Ka
22
When acetic acid ionizes in water,
following equilibrium is established:
+
→
HC H O (aq)
H +C H O
3
→
2
2
2
3
the
22
Ka is the ionization constant for this equilibrium.
+
2
[H ][C2H3O ]
Ka =
[HC2H3O2 ]
Ka is called the acid ionization constant.
Since the concentration of water is large and does
not change appreciably, it is omitted from Ka. 23
At 25oC, a 0.100 M solution of HC2H3O2 is 1.34%
ionized and has an [H+] of 1.34 x 10-3 mol/L.
Calculate Ka for acetic acid.
+
→
HC2H3O2(aq)
H + C2H3O22
→
Because each molecule of HC2H3O2 that
+
ionizes yields one H and one C2H33O22 , the
concentrations of the two ions are equal.
+
-3
[H ] = [C2H3O2 ] = 1.34 x 10 mol/L
The moles of unionized acetic acid per liter are
0.100 mol/L – 0.00134 mol/L = 0.099 mol/L
24
Substitute these concentrations into the
equilibrium expression and solve for Ka.
+
-3
[H ] = [C2H3O2 ] = 1.34 x 10 mol/L
[HC2H3O2] = 0.099 mol/L
+
2
[H ][C2H3O ]
Ka =
[HC2H3O2 ]
-3
-3
(1.34x10 )(1.34x10 )
-5

= 1.8x10
(0.099)
25
What is the [H+] in a 0.50 M HC2H3O2 solution? The
ionization constant, Ka, for HC2H3O2 is 1.8 x 10-5.
The equilibrium expression and Ka for HC2H3O2
are
+
→
HC2H3O2(aq)
H + C2H3O22
→
+
2
[H ][C2H3O ]
Ka =
[HC2H3O2 ]
+
[H ] = [C
O ] LetofY HC
Because
each
that
= [H
=2[C
2H3] O
2H3molecule
2H3O2 ]
+
ionizes
one
Hequilibrium
and one C2is
H330.50
O22 , the
[HCyields
H
O
]
at
– Y.
2 3 2
+
2
concentrations of the two ions are equal.
26
Substitute these values into Ka for HC2H3O2.
Y=
[H+]
2
= [C2H3O ] HC2H3O2 = 0.50 - Y
+
2
[H ][C2H3O ]
Ka =
[HC2H3O2 ]
2
2
2
(Y)(Y)
(Y)(Y) Y Y Y
-5 -5
-5
x 10
Ka =
= = = 1.8
= 1.8
=x1.8
10 x 10
0.50
0.50- 0.50
Y 0.50
0.50 - Y
Solve for Y2.
Assume Y is small
compared to
2
Then 0.50 – Y  0.50
0.50 -Y.
-5
Y = 0.50 x 1.8 x 10
27
Take the square root of both sides of the
2
-5
-5
-6
equation. Y = 0.50
100 x 10
0.90 x 1.8
10 x= 9.
-6
Y = 9.0 x 10
+
-3
= 3.0 x 10
[H ] = 3.0 x 10
-3
mol/L
mol/L
Making no approximation and using the
quadratic equation the answer is 2.99 x 10-3
mol/L, showing that it was justified to assume
Y was small compared to 0.5.
28
Calculate the percent ionization in a 0.50 M HC2H3O2
solution.
The ionization of a weak acid is given by
HA → H+ + A-
→
The percent ionization is given by
+
-
concentration of [H ] or [A ]
x 100= percent ionized
initial concentration of [HA]
30
Calculate the percent ionization in a 0.50 M HC2H3O2
solution.
The ionization of acetic acid is given by
+
→
HC2H3O2(aq)
H + C2H33O22
→
The percent ionization of acetic acid is given by
concentration of [H+ ] or [C2H3O-2 ]
x 100= percent ionized
initial concentration of [HC2H3O2 ]
-3 +
] was
3.0 x 10[H
mol
/ L previously calculated
x 100=
0.60% percent ionized
-3
as 3.0 x 10 mol/L
0.50 mol/L
31
Ionization Constants (Ka) of
Weak Acids at 25oC
Acid
Formula
Ka
Acid
Formula
Ka
Acetic
HC2H3O2
1.8 x 10-5
Hydrocyanic
HCN
4.0 x10-10
Benzoic
HC7H5O2
6.3 x 10-5
Hypochlorous
HOCl
3.5 x10-8
Carbolic
HC6H5O
1.3 x 10-10
Nitrous
HNO2
4.5 x10-4
Cyanic
HCNO
2.0 x 10-4
Hydrofluoric
HF
6.5 x10-4
Formic
HCHO2
1.8 x 10-4
32
Ionic Equilibria
Problems
33
Use Ka as a measure of acid strength.
Let 1 = Hi and 5 = Lo
Acid
Ka
order
HC2H3O2 1.8  10–5
H2S
6  10–8
HF
7  10– 4
HC2Cl3O2 2  10–1
HCN
7.2  10–10
35
Use Ka as a measure of acid strength.
Let 1 = Hi and 5 = Lo
Acid
Ka
HC2H3O2 1.8  10–5
order
3
H2S
6  10–8
4
HF
7  10– 4
2
HC2Cl3O2 2  10–1
1
HCN
7.2  10–10
5
36
Find the pH in 0.1 M HCN if
Ka = 7.2  10–10
Step 1: ionize the acid
→
HCN
→

H  CN
+
Step 2: write the Ka expression
+
-
[H ][CN ]
Ka =
[HCN]
37
Step 3: let x = [H+] = [CN–]
[HCN] = 0.1 M – x ≈ 0.1 M
Step 4: substitute into Ka expression
[x][x]
10
Ka =
 7.2  10
[0.1]
Step 5: solve for x:
x = 7.2 10
2
11
x  7.2 10
11
 8.5  10
6
38
Step 6: solve for pH
6
pH  - log 8.5 10  5.07
39
Find [H+] in 0.1 M H2CO3 if
Ka = 4.0  10–17
Step 1: ionize the acid
→
H2CO3
→
2H  CO
+
-2
3
Step 2: write the Ka expression
+ 2
-2
3
[H ] [CO ]
Ka =
[H2CO3 ]
40
Step 3: let x = [H+] , [CO3–2] = ½ x
[H2CO3] = 0.1 M – ½ x ≈ 0.1 M
Step 4: substitute into Ka expression
2 x
2
3
[x] [ ] x
17
Ka =

 4.0 10
[0.1] 0.2
Step 5: solve for x:
x = 8.0 10
3
18
x  8.0  10
3
18
 2.0  10
6
41
Find the pH of a 0.001 M solution of
bombastic hydroxide (BmOH) if
Kb = 1.6  10–10
Step 1: ionize the base
→
BmOH
→

Bm  OH
+
Step 2: write the Kb expression
+
-
[Bm ][OH ]
Kb =
[BmOH]
42
Step 3: let x = [Bm+] = [OH–]
[BmOH] = 0.001 M – x ≈ 0.001 M
Step 4: substitute into Kb expression
[x][x]
10
Kb =
 1.6  10
[0.001]
Step 5: solve for x:
x = 1.6 10
2
13
x  1.6 10
13
 4.0 10
7
43
Step 6: solve for pOH
7
pOH  - log 4.0 10  6.40
Step 7: solve for pH
pH  14  6.40  7.60
44
Find the Ka for a 0.01 M HCN
solution if the pH = 6.3
Step 1: ionize the acid
→
HCN
→

H  CN
+
Step 2: write the Ka expression
+
-
[H ][CN ]
Ka =
[HCN]
45
Step 3: use pH to find [H+]
[H+] = 10–pH = 10–6.3 = 5.0  10–7
[H+] = [CN–],
[HCN] = 0.01- 5.0  10–7 ≈ 0.01
Step 4: substitute into Ka expression
7
7
[5.0 10 ][5.0 10 ]
11
Ka =
 2.5 10
[0.01]
46
Solubility Product Constant
47
The solubility product constant, Ksp, is
the equilibrium constant of a slightly
(sparingly) soluble salt.
48
Silver chloride is in equilibrium with its ions in
aqueous solution.
AgCl(s) → Ag+(aq) + Cl-(aq)
→
The equilibrium constant+is
--
K sp = [Ag
][Cl
]]
[Ag
][Cl
The product of
Keq=and
K eq
[AgCl(s) is a constant.[AgCl(s )]
+
Rearrange
+
-
K eq x [AgCl(s)] = [Ag ][Cl ] = K sp
The The
amount
concentration
of solid
AgClof does
solid not
AgClaffect
is a
the equilibrium.
constant.
49
Ksp for Sparingly Soluble Solids
To find Ksp:
1. ionize the salt
2. write the Ksp expression
3. determine the M’s of the ions
4. substitute into the Ksp expression
50
The solubility of AgCl in water is 1.3 x 10-5
mol/L.
AgCl(s) → Ag+(aq) + Cl-(aq)
Because each formula unit of AgCl that
dissolves yields one Ag+ and one Cl- , the
concentrations of the two ions are equal.
→
[Ag+] = [Cl-] = 1.3 x 10-5 mol/L
Ksp = [Ag+][Cl-] The Ksp has no denominator.
= (1.3 x 10-5)(1.3 x 10-5) = 1.7 x 10-10
51
Molar Solubility for Sparingly
Soluble Solids
1.
2.
3.
4.
5.
6.
ionize the salt
write the Ksp expression
let y equal the amount of salt that
dissolves
find concentrations of ions in terms of y
substitute in the Ksp expression
solve for y
53
The Ksp value for lead sulfate is 1.3 x 10-8. Calculate
the solubility of PbSO4 in moles per liter.
The equilibrium equation of PbSO4 is
2+
2→
PbSO4
Pb + SO4
→
The Ksp of PbSO4 is
2+
2K sp = [Pb ][SO4 ]
Because each formula unit of PbSO4 that
22+
dissolves yields one Pb and one SO4 , the
concentrations of the two ions are equal.
2+
24
Let Y = [Pb ] = [SO ]
54
The Ksp value for lead sulfate is 1.3 x 10-8. Calculate
the solubility of PbSO4 in moles per liter.
Substitute Y into the Ksp equation.
2+
24
K sp = [Pb ][SO ]
K sp = (Y)(Y) = 1.3 x 10
2
-8
-8
Y = 1.3 x 10
-4
Y = 1.3 x 10 = 1.1 x 10 mol/L
-8
The solubility of PbSO4 is 1.1 x 10-4 mol/L.
55
Solubility Product Constants (Ksp) at 25oC
Compound
Ksp
Compound
Ksp
AgCl
1.7 x 10-10
CaF2
3.9 x 10-11
AgBr
5 x 10-13
CuS
9 x 10-45
AgI
8.5 x 10-17
Fe(OH)3
6 x 10-38
AgC2H3O2
2 x 10-3
PbS
7x 10-29
Ag2CrO4
1.9 x 10-12
PbSO4
1.3 x 10-8
BaCrO4
8.5 x 10-11
Mn(OH)2
2.0 x 10-13
BaSO4
1.5 x 10-9
56
Practice
57
The Ksp value for silver carbonate is
6.2 x 10-12. Calculate the solubility of
Ag2CO3 in moles per liter.
+
2→
Ag2CO3
2Ag + CO3
→
+ 2
23
Ksp = [Ag ] [CO ]
If y = amount of salt that dissolves,
then [Ag+] = 2y, [CO3–2] = y
58
2
K sp = [2y] [y] = 4y3
4y3 = 6.2 x 10-12
y3 = 1.6 x 10-12
3
y = 1.6 x 10
-12
y = 1.2 x 10-4 mol/L
The solubility of silver carbonate is
1.2 x 10-4 mol/L
59
Acid-Base
Properties of Salts
60
hydrolysis is the term used for the general
reaction in which a water molecule is split.
61
Salts that contain the anion of a weak acid
undergo hydrolysis.
The net ionic equation for the hydrolysis of
sodium acetate is
→
C2H3O (aq) + HOH(l) HC2H3O2 (aq) + OH (aq)
→
2
The water
The solution
molecule is
splits.
basic.
62
Salts that contain the cation of a weak
base undergo hydrolysis.
The net ionic equation for the hydrolysis of
ammonium chloride is
+
→
NH (aq) + HOH(l ) NH3 (aq) + H3O (aq)
→
+
4
The
Thesolution
water
molecule
is acidic.
splits.
63
Salts derived from a strong acid and a strong
base do not undergo hydrolysis.
Na+ + Cl- + HOH(l ) → no reaction
64
Ionic Composition of Salts and the Nature of
the Aqueous Solutions They Form
Type of salt
Nature of
Aqueous Solution
Examples
Weak base-strong acid
Acid
NH4Cl, NH4NO3
Strong base-weak acid
Basic
NaC2H3O2
Weak base-weak acid
Depends on the salt
NH4C2H3O2, NH4NO2
Strong base-strong acid
Neutral
NaCl, KBr
65
Buffer Solutions:
The Control of pH
66
A buffer solution resists changes in pH
when diluted or when small amounts of
acid or base are added.
67
A weak acid mixed with its conjugate base
form a buffer solution.
Sodium acetate when mixed with acetic acid
forms a buffer solution.
68
A weak acid mixed with its conjugate base
form a buffer solution.
If a small amount of HCl is added, the acetate
ions of the buffer will react with the H+ of the
HCl to form unionized acetic acid.
H (aq) + C2H3O (aq)  HC2H3O2 (aq)
+
2
69
A weak acid mixed with its conjugate base
form a buffer solution.
If a small amount of NaOH is added, the acetic
acid molecules of the buffer will react with the
OH– of the NaOH to form water.
OH + HC2H3O2 (aq)  H2O(l ) + C2H3O (aq)
-
2
70
A liter of a buffered solution
contains 0.1 moles of H2CO3
and 0.05 moles of NaHCO3.
Find the pH if Ka for the acid
is 1.0  10–7
71
72