1 chapter 16 ionic equilibria: acids and bases. 2 chapter goals 1. a review of strong electrolytes...
TRANSCRIPT
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1
Chapter 16
Ionic Equilibria: Acids and Bases
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2
Chapter Goals
1. A Review of Strong Electrolytes
2. The Autoionization of Water
3. The pH and pOH Scales
4. Ionization Constants for Weak Monoprotic Acids and Bases
5. Polyprotic Acids
6. Solvolysis
7. Salts of Strong Bases and Strong Acids
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3
Chapter Goals
8. Salts of Strong Bases and Weak Acids
9. Salts of Weak Bases and Strong Acids
10. Salts of Weak Bases and Weak Acids
11. Salts That Contain Small, Highly Charged Cations
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4
A Review of Strong Electrolytes
This chapter details the equilibria of weak acids and bases. We must distinguish weak acids and bases from strong
electrolytes. Weak acids and bases ionize or dissociate
partially, much less than 100%. In this chapter we will see that it is often less than 10%!
Strong electrolytes ionize or dissociate completely. Strong electrolytes approach 100% dissociation in
aqueous solutions.
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5
A Review of Strong Electrolytes There are three classes of strong electrolytes.1 Strong Water Soluble Acids
Remember the list of strong acids from Chapter 4.
3(aq)(aq)%100
)3(
3(aq)(aq)3100%
)(2)3(
NOHHNO
or
NOOH OHHNO
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A Review of Strong Electrolytes
3(aq)(aq)%100
)3(
3(aq)(aq)3100%
)(2)3(
NOHHNO
or
NOOH OHHNO
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7
A Review of Strong Electrolytes
2 Strong Water Soluble BasesThe entire list of these bases was also introduced in
Chapter 4.
-(aq)
2(aq)
100% OH2(s)
-(aq)(aq)
100% OH(s)
OH 2Sr Sr(OH)
OHKKOH
2
2
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8
A Review of Strong Electrolytes
3 Most Water Soluble SaltsThe solubility guidelines from Chapter 4 will help you remember
these salts.
3(aq)2(aq)
100% OHs23
-(aq)(aq)
100% OH(s)
NO 2Ca )Ca(NO
ClNaNaCl
2
2
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A Review of Strong Electrolytes The calculation of ion concentrations in solutions of
strong electrolytes is easy. Example 18-1: Calculate the concentrations of ions
in 0.050 M nitric acid, HNO3.
0.050 0.050 050.0
NOOHOHHNO 3(aq)(aq)3100%
)(2)3(
MMM
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10
A Review of Strong Electrolytes
Example 18-2: Calculate the concentrations of ions in 0.020 M strontium hydroxide, Sr(OH)2, solution.
You do it!
M
MMM
0.040
0.0202 0.020 020.0
OH 2SrSr(OH) -(aq)
2(aq)
OH2(s)
2
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11
The Autoionization of Water
Pure water ionizes very slightly. The concentration of the ionized water is less than one-
millionth molar at room temperature.
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The Autoionization of Water
We can write the autoionization of water as a dissociation reaction similar to those previously done in this chapter.
Because the activity of pure water is 1, the equilibrium constant for this reaction is:
-(aq)(aq)3)(2)(2 OHOHOH OH
K H O OHc 3+
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The Autoionization of Water
Experimental measurements have determined that the concentration of each ion is 1.0 x 10-7 M at 25oC. Note that this is at 25oC, not every temperature!
We can determine the value of Kc from this information.
K H O OH
1.0 x 10 1.0 x 10
1.0 x10
c 3+
-7 -7
14
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The Autoionization of Water
This particular equilibrium constant is called the ion-product for water and given the symbol Kw. Kw is one of the recurring expressions for the remainder of
this chapter and Chapters 19 and 20.
K H O OH
1.0 x10
w 3+
14
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15
The Autoionization of Water
Example 18-3: Calculate the concentrations of H3O+ and OH- in 0.050 M HCl.
].[OH calculate tous allow willK and OH The
.0.050OH theThus
0.050 0.050 050.0
Cl OH OH + HCl
-w
+3
+3
+32
M
MMM
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16
The Autoionization of Water
Use the [H3O+] and Kw to determine the [OH-].
You do it!
M13
2
14
+3
14
14+3
100.2OH
100.5
100.1
OH
100.1OH
100.1OHOH
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The Autoionization of Water
The increase in [H3O+] from HCl shifts the equilibrium and decreases the [OH-]. Remember from Chapter 17, increasing the product
concentration, [H3O+], causes the equilibrium to shift to the reactant side.
This will decrease the [OH-] because it is a product!
. 0.050 10 2.0 0.050 ]O[H overall The
.102.0 is K from ]O[H The
OH OH OH OH
0.050 is HCl from ]O[H The
13-3
13-w3
-322
3
MM
M
M
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The Autoionization of Water
Now that we know the [H3O+] we can calculate the [OH-].
You do it!You do it!
M
M
M
13-
14--
3
14--
3
-143w
100.2]OH[
] [0.050
101]OH[
]O[H
101]OH[
. 0.050 ]O[H Since
101][OH-]O[HK
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The pH and pOH scales
A convenient way to express the acidity and basicity of a solution is the pH and pOH scales.
The pH of an aqueous solution is defined as:
pH = -log H O3+
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The pH and pOH scales
In general, a lower case p before a symbol is read as the ‘negative logarithm of’ the symbol.
Thus we can write the following notations.
.quantitiesother for forth so and
Ag-log=pAg
OH-log=pOH+
-
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The pH and pOH scales
If either the [H3O+] or [OH-] is known, the pH and pOH can be calculated.
Example 18-4: Calculate the pH of a solution in which the [H3O+] =0.030 M.
52.1pH
100.3logpH
OH-log=pH2
+3
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The pH and pOH scales
Example 18-5: The pH of a solution is 4.597. What is the concentration of H3O+?
You do it!You do it!
M53
4.597-3
3
3
3
1053.2]O[H
10]O[H
-4.597]Olog[H
]O-log[H4.597
]O-log[HpH
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The pH and pOH scales
A convenient relationship between pH and pOH may be derived for all dilute aqueous solutions at 250C.
Taking the logarithm of both sides of this equation gives:
143 100.1]][OHO[H
00.14OHlogOHlog 3
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The pH and pOH scales
Multiplying both sides of this equation by -1 gives:
Which can be rearranged to this form:
00.14OHlogOHlog- 3
14.00 pOH pH
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The pH and pOH scales
Remember these two expressions!! They are key to the next three chapters!
14.00pOH pH
100.1OH OH 143
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The pH and pOH scales
The usual range for the pH scale is 0 to 14.
And for pOH the scale is also 0 to 14 but inverted from pH. pH = 0 has a pOH = 14 and pH = 14 has a pOH = 0.
14.00pH to0pH
100.1OH to 0.1OH 1433
MM
0pOH 00.14pOH
0.1OH toup 100.1OH 14
MM
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The pH and pOH scales
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The pH and pOH scales
Example 18-6: Calculate the [H3O+], pH, [OH-], and pOH for a 0.020 M HNO3 solution. Is HNO3 a weak or strong acid?
What is the [H3O+] ?
70.1pH
100.2-logpH
100.2OH
0.020 0.020 020.0
NOOHOHHNO
2
23
-33
100%23
M
M
MMM
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The pH and pOH scales
Example 18-6: Calculate the [H3O+], pH, [OH-], and pOH for a 0.020 M HNO3 solution.
30.12100.5logpOH
100.5100.2
100.1
OH
100.1OH
100.1OHOHK
13
132
14
3
14
143w
M
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The pH and pOH scales To help develop familiarity with the pH and pOH scale we can
look at a series of solutions in which [H3O+] varies between 1.0 M and 1.0 x 10-14 M.
[H3O+] [OH-] pH pOH
1.0 M 1.0 x 10-14 M 0.00 14.00
1.0 x 10-3 M 1.0 x 10-11 M 3.00 11.00
1.0 x 10-7 M 1.0 x 10-7 M 7.00 7.00
2.0 x 10-12 M 5.0 x 10-3 M 11.70 2.30
1.0 x 10-14 M 1.0 M 14.00 0.00
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Ionization Constants for Weak Monoprotic Acids and Bases Let’s look at the dissolution of acetic acid, a weak
acid, in water as an example. The equation for the ionization of acetic acid is:
The equilibrium constant for this ionization is expressed as:
-3323 COOCHOHOH COOHCH
OH COOHCH
COOCH OHK
23
33c
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Ionization Constants for Weak Monoprotic Acids and Bases The water concentration in dilute aqueous solutions
is very high. 1 L of water contains 55.5 moles of water. Thus in dilute aqueous solutions:
M5.55OH2
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Ionization Constants for Weak Monoprotic Acids and Bases The water concentration is many orders of
magnitude greater than the ion concentrations. Thus the water concentration is essentially that of
pure water. Recall that the activity of pure water is 1.
COOHCH
COOCH OH K
COOHCH
COOCH OHOHK
3
33
3
332c
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Ionization Constants for Weak Monoprotic Acids and Bases We can define a new equilibrium constant for weak
acid equilibria that uses the previous definition. This equilibrium constant is called the acid ionization
constant. The symbol for the ionization constant is Ka.
acid aceticfor
108.1COOHCH
COOCH OHK 5
3
33a
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Ionization Constants for Weak Monoprotic Acids and Bases In simplified formsimplified form the dissociation equation and
acid ionization expression are written as:
5
3
3a
-33
108.1COOHCH
COOCH HK
COOCHHCOOHCH
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Ionization Constants for Weak Monoprotic Acids and Bases The ionization constant values for several acids are
given below. Which acid is the strongest?
Acid Formula Ka value
Acetic CH3COOH 1.8 x 10-5
Nitrous HNO2 4.5 x 10-4
Hydrofluoric HF 7.2 x 10-4
Hypochlorous HClO 3.5 x 10-8
Hydrocyanic HCN 4.0 x 10-10
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Ionization Constants for Weak Monoprotic Acids and Bases From the above table we see that the order of
increasing acid strength for these weak acids is:
The order of increasing base strength of the anions (conjugate bases) of these acids is:
HCN>HClO>COOHCH>HNO>HF 32
---3
-2
- CN<ClO<COOCH<NO<F
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Ionization Constants for Weak Monoprotic Acids and Bases Example 18-8: Write the equation for the ionization
of the weak acid HCN and the expression for its ionization constant.
10--
a
-
10 x 4.0HCN
CN HK
CN H HCN
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Ionization Constants for Weak Monoprotic Acids and Bases Example 18-9: In a 0.12 M solution of a weak
monoprotic acid, HY, the acid is 5.0% ionized. Calculate the ionization constant for the weak acid.
You do it!You do it!
HY H + Y
KH Y
HY
+ -
a
+ -
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Ionization Constants for Weak Monoprotic Acids and Bases Since the weak acid is 5.0% ionized, it is also 95% unionized. Calculate the concentration of all species in solution.
MM
M
MM
11.0)12.0(95.0HY
100.6YH
0060.0)12.0(05.0YH3+
+
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Ionization Constants for Weak Monoprotic Acids and Bases
Use the concentrations that were just determined in the ionization constant expression to get the value of Ka.
4
a
33
a
a
103.3K
11.0
100.6 100.6K
HY
Y HK
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Ionization Constants for Weak Monoprotic Acids and Bases Example 18-10: The pH of a 0.10 M solution of a
weak monoprotic acid, HA, is found to be 2.97. What is the value for its ionization constant?
pH = 2.97 so [H+]= 10-pH
M3
97.2
101.1H
10H
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Ionization Constants for Weak Monoprotic Acids and Bases Use the [H3O+] and the ionization reaction to
determine concentrations of all species.
HA H A
Equil. []'s 0.10 -1.1 10 1.1 10 1.1 10
0.10
+ -
-3 -3 -3
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Ionization Constants for Weak Monoprotic Acids and Bases Calculate the ionization constant from this
information.
5a
-3-3
a
102.1K
0.10
101.1101.1
HA
AHK
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46
Ionization Constants for Weak Monoprotic Acids and Bases Example 18-11: Calculate the concentrations of the
various species in 0.15 M acetic acid, CH3COOH, solution.
It is always a good idea to write down the ionization reaction and the ionization constant expression.
5
3
-33
a
-3323
108.1COOHCH
COOCHOHK
COOCHOH OHCOOHCH
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47
Ionization Constants for Weak Monoprotic Acids and Bases Next, combine the basic chemical concepts with
some algebra to solve the problem.
M0.15 [] Initial
COOCH OH OHCOOHCH -3323
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48
Ionization Constants for Weak Monoprotic Acids and Bases Next we combine the basic chemical concepts with
some algebra to solve the problem
xMxMxM
M
- Change
0.15 [] Initial
COOCH OH OHCOOHCH -3323
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49
Ionization Constants for Weak Monoprotic Acids and Bases Next we combine the basic chemical concepts with
some algebra to solve the problem
xMxM-x)M.(
xMxMxM
M
150 [] mEquilibriu
- Change
0.15 [] Initial
COOCH OH OHCOOHCH -3323
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50
Substitute these algebraic quantities into the ionization expression.
5
3
33a
108.115.0
COOHCH
COOCH OHK
x
xx
Ionization Constants for Weak Monoprotic Acids and Bases
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51
Ionization Constants for Weak Monoprotic Acids and Bases Solve the algebraic equation, using a simplifying assumption that is
appropriate for all weak acid and base ionizations.
52
52
108.115.0
108.115.0
xx
x
x
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52
Solve the algebraic equation, using a simplifying assumption that is appropriate for all weak acid and base ionizations.
Ionization Constants for Weak Monoprotic Acids and Bases
52
3a
52
52
108.115.0
[]. tocompared ignore enough to small is
.assumption thismake then 10 K If
108.115.0
108.115.0
x
x
xx
x
x
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53
Complete the algebra and solve for the concentrations of the species.
Ionization Constants for Weak Monoprotic Acids and Bases
MM
Mx
x
15.0106.115.0COOHCH
COOCHOH106.1
107.2
33
333
62
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54
Ionization Constants for Weak Monoprotic Acids and Bases Note that the properly applied simplifying assumption gives
the same result as solving the quadratic equation does.
2a
4acbb
c b a
0107.2108.1
108.115.0
2
652
5
x
xx
X
xx
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55
Ionization Constants for Weak Monoprotic Acids and Bases
3-3
6255
101.6- and 106.1
12
107.214108.1108.1
x
x
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56
Ionization Constants for Weak Monoprotic Acids and Bases Let us now calculate the percent ionization for the
0.15 M acetic acid. From Example 18-11, we know the concentration of CH3COOH that ionizes in this solution. The percent ionization of acetic acid is
%1.1%10015.0
106.1ionization %
%100COOHCH
COOHCH= ionization %
3
original3
ionized3
M
M
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57
Ionization Constants for Weak Monoprotic Acids and Bases Example 18-12: Calculate the concentrations
of the species in 0.15 M hydrocyanic acid, HCN, solution.
Ka= 4.0 x 10-10 for HCN
You do it!You do it!
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58
Ionization Constants for Weak Monoprotic Acids and Bases
MMx
Mx
x
x
xx
MxMxMx
MxMxMx
M
15.0 15.0HCN
CNH107.7
100.6
100.415.0
HCN
CN HK
-0.15 mEquilibriu
+ + - Change
0.15 Initial
CN OH OH HCN
6
112
10a
-32
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59
Ionization Constants for Weak Monoprotic Acids and Bases The percent ionization of 0.15 M HCN solution is
calculated as in the previous example.
%0051.0%10015.0
107.7ionization %
%100HCN
HCN = ionization %
6
original
ionized
M
M
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60
Ionization Constants for Weak Monoprotic Acids and Bases Let’s look at the percent ionization of two weak acids
as a function of their ionization constants. Examples 18-11 and 18-12 will suffice.
Note that the [H+] in 0.15 M acetic acid is 210 times greater than for 0.15 M HCN.
Solution Ka [H+] pH % ionization
0.15 M acetic acid
1.8 x 10-5 1.6 x 10-3 2.80 1.1
0.15 M
HCN
4.0 x 10-10 7.7 x 10-6 5.11 0.0051
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61
Ionization Constants for Weak Monoprotic Acids and Bases All of the calculations and understanding we have at
present can be applied to weak acids and weak bases!
One example of a weak base ionization is ammonia ionizing in water.
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62
Ionization Constants for Weak Monoprotic Acids and Bases All of the calculations and understanding we have at
present can be applied to weak acids and weak bases!
Example 18-13: Calculate the concentrations of the various species in 0.15 M aqueous ammonia.
xMxM-x)M.(
xMxMxM
M
150 [] mEquilibriu
- Change
0.15 [] Initial
OH NH OH NH -423
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63
Ionization Constants for Weak Monoprotic Acids and Bases
MM
M
Mxx
x
xx
x
xx
x
xx
xMxM-x)M.(
xMxMxM
M
15.0101.615.0NH
101.6OHNH
101.6 and 107.2
108.1)15.0(
108.115.0
15.0
0.15 x - 0.15 thus0.15x
valid.is assumption gsimplifyin The
108.115.0
NH
OH NHK
150 [] mEquilibriu
- Change
0.15 [] Initial
OH NH OH NH
3-3
3--4
3-62
52
5
5
3
-4
b
-423
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64
Ionization Constants for Weak Monoprotic Acids and Bases The percent ionization for weak bases is calculated
exactly as for weak acids.
%1.1
%10015.0
106.1
%100NH
NHionization %
3
original3
ionized3
M
M
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65
Ionization Constants for Weak Monoprotic Acids and Bases Example 18-14: The pH of an aqueous
ammonia solution is 11.37. Calculate the molarity (original concentration) of the aqueous ammonia solution.
You do it!You do it!
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66
Ionization Constants for Weak Monoprotic Acids and Bases
M
M3
4
363.2pOH-
103.2NH
103.21010OH
2.63 11.37-14.00pH - 14.00=pOH
pOH. thederivecan we14.00, = pOH + pH From
11.37=pH
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67
Ionization Constants for Weak Monoprotic Acids and Bases Use the ionization equation and some algebra to get
the equilibrium concentration.
3-3-3-
3-3-3-
-423
102.3+ 102.3+ 102.3- m[]Equilibriu
102.3+ 102.3+ 102.3- Change
Initial[]
OH NH OH NH
Mx
xM
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68
Ionization Constants for Weak Monoprotic Acids and Bases Substitute these values into the ionization constant
expression.
3
335
5
3
4b
103.2
103.2 103.2108.1
108.1NH
OH NHK
x
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69
Ionization Constants for Weak Monoprotic Acids and Bases Examination of the last equation suggests that our
simplifying assumption can be applied. In other words (x-2.3x10-3) x.
Making this assumption simplifies the calculation.
3
5
23
NH 30.0
108.1103.2
Mxx
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70
Polyprotic Acids Many weak acids contain two or more acidic hydrogens.
Examples include H3PO4 and H3AsO4.
The calculation of equilibria for polyprotic acids is done in a stepwise fashion. There is an ionization constant for each step.
Consider arsenic acid, H3AsO4, which has three ionization constants.
1 Ka1 = 2.5 x 10-4
2 Ka2 = 5.6 x 10-8
3 Ka3 = 3.0 x 10-13
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71
Polyprotic Acids
The first ionization step for arsenic acid is:
4
43
421
-4243
105.2AsOH
AsOH HKa
AsOHHAsOH
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72
Polyprotic Acids
The second ionization step for arsenic acid is:
8-1
42
24
a2
-24
-142
106.5AsOH
HAsO HK
HAsOHAsOH
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73
Polyprotic Acids
The third ionization step for arsenic acid is:
13-2
4
34
a3
-34
-24
100.3HAsO
AsOHK
AsO HHAsO
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74
Polyprotic Acids
Notice that the ionization constants vary in the following fashion:
This is a general relationship. For weak polyprotic acids the Ka1 is always > Ka2, etc.
a3a2a1 KKK
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75
Polyprotic Acids
Example 18-15: Calculate the concentration of all species in 0.100 M arsenic acid, H3AsO4, solution.
1 Write the first ionization step and represent the concentrations.Approach this problem exactly as previously done.
xMxMMx 100.0
AsOHHAsOH 4243
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76
Polyprotic Acids
2 Substitute the algebraic quantities into the expression for Ka1.
apply.not does assumption gsimplifyin thecase, In this
0105.2105.2
105.210.0
K
105.2AsOH
AsOH HK
542
4a1
4
43
42a1
xx
x
xx
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77
Polyprotic Acids
Use the quadratic equation to solve for x, and obtain both values of x.
MMx
MxM
MxMx
x
095.0100.0AsOH
109.4AsOHH
109.4 and 101.5
12
105.214105.2105.2
43
342
33
5244
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78
Polyprotic Acids
4 Next, write the equation for the second step ionization and represent the concentrations.
yMyMMy
M
)10(4.9lly algebraica
)10(4.9 step1st ]from [
HAsO + H AsOH
3-
3-
-24
+-42
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79
Polyprotic Acids
5 Substitute the algebraic expressions into the second step ionization expression.
.assumption apply thecan westep For this
104.9
104.9=K
106.5AsOH
HAsO OH=K
3-
3-
a2
8
42
243
a2
y
yy
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80
Polyprotic Acids
2nd1st
242nd
8
83-
3-
a2
3-3-
3-
3-
3-
a2
8
42
243
a2
HH that Note
HAsOH106.5
106.5104.9
104.9=K
104.9104.9 Thus,
104.9 y
applied. becan assumption thecase In this
104.9
104.9=K
106.5AsOH
HAsOOH=K
My
y
y
y
yy
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81
Polyprotic Acids
6 Finally, repeat the entire procedure for the third ionization step.
MzMzMz
MM
105.6 changes ] [ of tionsrepresenta algebraic
105.6104.9 105.6 sionization 2 and 1 from s]' [
AsO H HAsO
8-
8-3-8-ndst
-34
24
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82
Polyprotic Acids
7. Substitute the algebraic representations into the third ionization expression.
.105.6 z applied, becan assumption The
106.5
106.5109.4=K
100.3HAsO
AsOOH=K
8-
8
83
a3
1324
343
a3
z
zz
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83
Polyprotic Acids
Use Kw to calculate the [OH-] in the 0.100 M H3AsO4 solution.
M12
3
1414
14
100.2OH
109.4
100.1
H
100.1OH
100.1OHH
343rd
18
138
3
AsOH104.3
100.3106.5
109.4
Mz
z
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84
Polyprotic Acids A comparison of the various species in 0.100 M
H3AsO4 solution follows.
Species Concentration
H3AsO4 0.095 M
H+ 0.0049 M
H2AsO4- 0.0049 M
HAsO42- 5.6 x 10-8 M
AsO43- 3.4 x 10-18 M
OH- 2.0 x 10-12 M
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85
Solvolysis This reaction process is the most difficult concept in this chapter. Solvolysis is the reaction of a substance with the solvent in which
it is dissolved. Hydrolysis refers to the reaction of a substance with water or its
ions. Combination of the anion of a weak acid with H3O+ ions from
water to form nonionized weak acid molecules.
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86
Solvolysis Hydrolysis refers to the reaction of a
substance with water or its ions. Hydrolysis is solvolysis in aqueous solutions.
The combination of a weak acid’s anion with H3O+ ions, from water, to form nonionized weak acid molecules is a form of hydrolysis.
A H O HA H O
recall H O + H O H O OH
-3 2
2 2 3
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87
Solvolysis
The reaction of the anion of a weak monoprotic acid with water is commonly represented as:
mequilibriu water the
upsets OH of removal The
OH HA OH A+
3
-2
-
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88
Solvolysis
Recall that at 25oC in neutral solutions:
[H3O+] = 1.0 x 10-7 M = [OH-]
in basic solutions:[H3O+] < 1.0 x 10-7 M and [OH-] > 1.0 x 10-7 M
in acidic solutions:
[OH-] < 1.0 x 10-7 M and [H3O+] > 1.0 x 10-7 M
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89
Solvolysis
Remember from BrØnsted-Lowry acid-base theory: The conjugate base of a strong acid is a very weak base. The conjugate base of a weak acid is a stronger base. Hydrochloric acid, a typical strong acid, is essentially
completely ionized in dilute aqueous solutions.
HCl H O H O Cl2 3 ~100%
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90
Solvolysis The conjugate base of HCl, the Cl- ion, is a very weak base.
The chloride ion is such a weak base that it will not react with the hydronium ion.
This fact is true for all strong acids and their anions.
Cl H O No rxn. in dilute aqueous solutions3
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91
Solvolysis HF, a weak acid, is only slightly ionized in dilute aqueous solutions. Its conjugate base, the F- ion, is a much stronger base than the Cl-
ion. The F- ions combine with H3O+ ions to form nonionized HF.
Two competing equilibria are established.
HF + H O H O F
only slightly
F + H O HF + H O
nearly completely
2 3+ -
-3
+2
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92
Solvolysis Dilute aqueous solutions of salts that
contain no free acid or base come in four types:
1. Salts of Strong Bases and Strong Acids
2. Salts of Strong Bases and Weak Acids
3. Salts of Weak Bases and Strong Acids
4. Salts of Weak Bases and Weak Acids
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93
Salts of Strong Bases and Weak Acids Salts made from strong acids and strong soluble
bases form neutral aqueous solutions. An example is potassium nitrate, KNO3, made from
nitric acid and potassium hydroxide.
neutral. issolution theThus
OHOHupset oreaction t no is There
amounts. equalin present are HNO and KOH The
solution in are that ions The
OH OH OHOH
NOK KNO
-+3
3
HNOKOH
3-
22
3+OHin %100~
)(3
3
2
s
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94
Salts of Strong Bases and Weak Acids Salts made from strong soluble bases and weak acids hydrolyze to
form basic solutions.
Anions of weak acids (strong conjugate bases) react with water to form hydroxide ions.
An example is sodium hypochlorite, NaClO, made from sodium hydroxide and hypochlorous acid.
base?or acidstronger theisWhich
solution in ions Notice
OH + OH OH + OH
ClONaNaClO
HClONaOH
3-
22
-OHin %100~)(
2
s
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95
Salts of Strong Bases and Weak Acids
Na ClO Na ClO
H O + H O OH + H O
ClO H O HClO H O
+ - in H O -
2 2-
3+
-3 2
2( )
~s
100%
We can combine these last two equations into one single equation that represents the total reaction.
ClO H O HClO OH-2
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96
Salts of Strong Bases and Weak Acids
K =
HClO OH
ClOb
-
-
The equilibrium constant for this reaction, called the hydrolysis constant, is written as:
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97
Salts of Strong Bases and Weak Acids Algebraic manipulation of the previous expression
give us a very useful form of the expression. Multiply the expression by one written as [H+]/ [H+].
H+/H+ = 1
K =HClO OH
ClO
H
H
K =HClO
H ClO
H OH
b
-
-
b -
-
1
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98
Salts of Strong Bases and Weak Acids Which can be rewritten as:
K =
HClO
H ClO
H OH
K =1
KK
b -
-
ba for HClO
w
1
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99
Salts of Strong Bases and Weak Acids Which can be used to calculate the hydrolysis
constant for the hypochlorite ion:
K =1
KK
K =K
K=
1 10
3.5 10
K =HClO OH
ClO
ba for HClO
w
bw
a for HClO
-14
-8
b
2 9 10 7.
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100
Salts of Strong Bases and Weak Acids This same method can be applied to the anion of
any weak monoprotic acid.
HAfor a
Wb
-2
K
K
A
OHHA=K
OHHAOHA
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101
Salts of Strong Bases and Weak Acids Example 18-16: Calculate the hydrolysis constants
for the following anions of weak acids.
1. The fluoride ion, F-, the anion of hydrofluoric acid, HF. For HF, Ka=7.2 x 10-4.
F H O HF OH
KHF OH
F
KK
K
2
bw
a for HF
b
10 10
7 2 1014 10
14
411.
..
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102
Salts of Strong Bases and Weak Acids The cyanide ion, CN-, the anion of hydrocyanic acid, HCN.
For HCN, Ka = 4.0 x 10-10.
You do it!You do it!
510
14
b
HCNfor a
wb
-2
105.2100.4
100.1K
K
K
CN
OHHCNK
OH+ HCNOHCN
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103
Salts of Strong Bases and Weak Acids Example 18-17: Calculate [OH-], pH and percent
hydrolysis for the hypochlorite ion in 0.10 M sodium hypochlorite, NaClO, solution. “Clorox”, “Purex”, etc., are 5% sodium hypochlorite solutions.
MMM 0.10 0.10 0.10
ClO Na NaClO OHin 100%~(s)
2
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104
Salts of Strong Bases and Weak Acids Set up the equation for the hydrolysis and the
algebraic representations of the equilibrium concentrations.
ClO + H O HClO + OH
Initial: 0.10
Change: - + +
At equil: 0.10 -
-2
-M M M
xM xM xM
x M xM xM
0 0
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105
Salts of Strong Bases and Weak Acids Substitute the algebraic expressions into the
hydrolysis constant expression.
7b 109.2
ClO
OH HClOK
Kb
x xx010
2 9 10 7
..
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106
Salts of Strong Bases and Weak Acids Substitute the algebraic expressions into the
hydrolysis constant expression.
10.23. pH get the we14.00 pOH pH From
3.77 pOH get the we][OH theFrom
OHClO107.1 becomesWhich
109.2 toreducesequation The
0.10 -0.10 and 0.10
case. in this made becan assumption gsimplifyin The
-
4
82
Mx
x
xx
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107
Salts of Strong Bases and Weak Acids The percent hydrolysis for the hypochlorite ion may
be represented as:
%17.0%1000.10
101.7=hydrolysis %
%100ClO
ClO=hydrolysis %
4-
original-
hydrolyzed-
M
M
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108
Salts of Strong Bases and Weak Acids If a similar calculation is performed for 0.10 M NaF
solution and the results from 0.10 M sodium fluoride and 0.10 M sodium hypochlorite compared, the following table can be constructed.
Solution Ka Kb [OH-] (M) pH%
hydrolysis
NaF 7.2 x 10-4 1.4 x 10-11 1.2 x 10-6 8.08 0.0012
NaClO 3.5 x 10-8 2.9 x 10-7 1.7 x 10-4 10.23 0.17
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109
Salts of Weak Bases and Strong Acids Salts made from weak bases and strong acids form
acidic aqueous solutions. An example is ammonium bromide, NH4Br, made
from ammonia and hydrobromic acid.
base?or acidstronger theisWhich
aresolution in Ions
OH OH OHOH
Br NH BrNH
HBrOHNH
3-
22
-4
100%~OHs4
4
2
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110
Salts of Weak Bases and Strong Acids
OH excess generates
OHNHOHNH
OH excess leavingsolution fromit
removingion OH with thereacts
,NH acid, strong relatively The
3
23-
4
3
-
4
The reaction may be more simply represented as:
OHNH OHNH 3324
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111
Salts of Weak Bases and Strong Acids
The hydrolysis constant expression for this process is:
Or even more simply as:
4
3a
4
33a NH
H NHKor
NH
OH NHK
HNH NH 34
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112
Salts of Weak Bases and Strong Acids Multiplication of the hydrolysis constant expression
by [OH-]/ [OH-] gives:
1
OH OH
OH NH
NHK
OH
OH
NH
OH NHK
-3
-4
3a
-
-
4
33a
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113
Salts of Weak Bases and Strong Acids Which we recognize as:
K
KK1
KK
K
ab NH
w w
b NH
a
3 3
1
10 10
18 105 6 10
14
510.
..
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114
Salts of Weak Bases and Strong Acids In its simplest form for this hydrolysis:
NH NH H
KNH H
NH
4 3
a3
4
5 6 10 10.
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115
Salts of Weak Bases and Strong Acids Example 18-18: Calculate [H+], pH, and percent
hydrolysis for the ammonium ion in 0.10 M ammonium bromide, NH4Br, solution.
1. Write down the hydrolysis reaction and set up the table as we have done before:
NH H O NH H
Initial[]: 0.10 + +
Change: - + +
Equilibrium[]: 0.10
4+
2 3
M xM xM
xM xM xM
x M xM xM
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116
Salts of Weak Bases and Strong Acids2. Substitute the algebraic expressions into the
hydrolysis constant.
0.10 - 0.10 thus0.10
.applicable is assumption The
105.6=10.0
K
106.5NH
H NHK
10-a
10
4
3a
xx
x
xx
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117
Salts of Weak Bases and Strong Acids3. Complete the algebra and determine the
concentrations and pH.
12.5pH
107.5HNH
107.5=
106.5
106.510.0
6-3
6-
112
10
M
Mx
x
x
xx
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118
Salts of Weak Bases and Strong Acids4. The percent hydrolysis of the ammonium ion in
0.10 M NH4Br solution is:
%
%.
%
hydrolysis =NH
NH
hydrolysis =7.5 10
hydrolysis = 0.0075%
4+
hydrolized
4+
original
-6
100%
010100%
MM
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119
Salts of Weak Bases and Weak Acids Salts made from weak acids and weak bases can
form neutral, acidic or basic aqueous solutions. The pH of the solution depends on the relative values of
the ionization constant of the weak acids and bases.
1. Salts of weak bases and weak acids for which parent Kbase =Kacid make neutral solutions.
An example is ammonium acetate, NH4CH3COO, made from aqueous ammonia, NH3,and acetic acid, CH3COOH.
Ka for acetic acid = Kb for ammonia = 1.8 x 10-5.
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120
Salts of Weak Bases and Weak Acids The ammonium ion hydrolyzes to produce H+ ions.
Its hydrolysis constant is:
NH NH H
KNH H
NH
4+
3
a3
4+
5 6 10 10.
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121
Salts of Weak Bases and Weak Acids The acetate ion hydrolyzes to produce OH- ions. Its
hydrolysis constant is:
CH COO H O CH COOH OH
KCH COOH OH
CH COO
3 2 3
b3
3
5 6 10 10.
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122
Salts of Weak Bases and Weak Acids Because the hydrolysis constants for both ions are
equal, their aqueous solutions are neutral. Equal numbers of H+ and OH- ions are produced.
solution!in formed are base and acidA weak
aresolution in Ions
OH OH OH OH
COOCHNHCOOCHNH
COOHCHOHNH
3-
22
34100%~OH
34
34
2
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123
Salts of Weak Bases and Weak Acids2. Salts of weak bases and weak acids for
which parent Kbase > Kacid make basic solutions.
An example is ammonium hypochlorite, NH4ClO, made from aqueous ammonia, NH3,and hypochlorous acid, HClO.
Kb for NH3 = 1.8 x 10-5 > Ka for HClO = 3.5x10-8
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124
Salts of Weak Bases and Weak Acids The ammonium ion hydrolyzes to produce H+ ions.
Its hydrolysis constant is:
10
4
3a
34
106.5NH
H NHK
HNHNH
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125
Salts of Weak Bases and Weak Acids The hypochlorite ion hydrolyzes to produce OH-
ions. Its hydrolysis constant is:
Because the Kb for ClO- ions is three orders of magnitude larger than the Ka for NH4
+ ions, OH- ions are produced in excess making the solution basic.
ClO H O HClO OH
KHClO OH
ClO
2
b -
2 9 10 7.
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126
Salts of Weak Bases and Weak Acids3. Salts of weak bases and weak acids for which
parent Kbase < Kacid make acidic solutions. An example is trimethylammonium fluoride,
(CH3)3NHF, made from trimethylamine, (CH3)3N,and hydrofluoric acid acid, HF.
Kb for (CH3)3N = 7.4 x 10-5 < Ka for HF = 7.2 x 10-4
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127
Salts of Weak Bases and Weak Acids Both the cation, (CH3)3NH+, and the anion, F-,
hydrolyze.
CH NH F CH NH F3 3H O~100%
3 32
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128
Salts of Weak Bases and Weak Acids The trimethylammonium ion hydrolyzes to produce
H+ ions. Its hydrolysis constant is:
(CH ) NH CH N H
KCH N H
(CH ) NH
KK
K
3 3+
3
a3
3 3+
w
b for CH N
a
3
( )
( )
.
..
( )
3
3
14
510
3
10 10
7 4 1014 10
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129
Salts of Weak Bases and Weak Acids The fluoride ion hydrolyzes to produce OH- ions. Its
hydrolysis constant is:
Because the Ka for (CH3)3NH+ ions is one order of magnitude larger than the Kb for F- ions, H+ ions are produced in excess making the solution acidic.
F H O HF OH
KHF OH
F
KK
K
2
b -w
a for HF
b
10 10
7 2 1014 10
14
411.
..
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Salts of Weak Bases and Weak Acids Summary of the major points of
hydrolysis up to now.1 The reactions of anions of weak monoprotic
acids (from a salt) with water to form free molecular acids and OH-.
A + H O HA + OH
KK
K
-2
-
bw
a HA
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Salts of Weak Bases and Weak Acids2. The reactions of anions of weak monoprotic acids
(from a salt) with water to form free molecular acids and OH-.
BH + H O B + H O
KK
K B = weak base
+2 3
+
aw
b B
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Salts of Weak Bases and Weak Acids Aqueous solutions of salts of strong acids
and strong bases are neutral. Aqueous solutions of salts of strong bases
and weak acids are basic. Aqueous solutions of salts of weak bases and
strong acids are acidic. Aqueous solutions of salts of weak bases and
weak acids can be neutral, basic or acidic.The values of Ka and Kb determine the pH.
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Hydrolysis of Small Highly-Charged Cations Cations of insoluble bases (metal hydroxides) become
hydrated in solution. An example is a solution of Be(NO3)3. Be2+ ions are thought to be tetrahydrated and sp3 hybridized.
Be Be(OH )
1s 2s 2p
Be
aq2+
2 4
4
2
1s 2s 2p
Be
Be(OH )
e pairs on coordinated water molecules
2
form sp hybrids
2 4
-
3
2 xx xx xx xx
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Hydrolysis of Small Highly-Charged Cations In condensed form it is represented as:
or, even more simply as:
Be(OH ) H O Be OH OH H O2 4 2 2 3 32
Be H O Be(OH) H2+2
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Hydrolysis of Small Highly-Charged Cations The hydrolysis constant expression for [Be(OH2)4]2+
and its value are:
or, more simply
K
Be(OH ) (OH) H O
Be(OH )a2 3
+3
2 42
10 10 5.
K
Be(OH) H
Bea
+
2+
10 10 5.
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Hydrolysis of Small Highly-Charged Cations Example 18-19: Calculate the pH and percent
hydrolysis in 0.10 M aqueous Be(NO3)2 solution.
1. The equation for the hydrolysis reaction and representations of concentrations of various species are:
xMxMMx 10.0
H Be(OH) OH Be 22
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Hydrolysis of Small Highly-Charged Cations2. Algebraic substitution of the expressions into the
hydrolysis constant:
00.3pH
100.1BeH
100.1
100.1
100.110.0
10.0 - 0.10 and 0.10
applies. assumption gsimplifyin The
100.110.0
3hydrolyzed
2
3
62
5
5
M
Mx
x
xx
xx
x
xx
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Hydrolysis of Small Highly-Charged Cations3 Calculate the percent hydrolysis of Be2+.
% hydrolyzed =Be
Be
% hydrolyzed =1.0 10
0.10
2+
hydrolyzed2+
original
-3
100%
100% 10%.
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Hydrolysis of Small Highly-Charged Cations This table is a comparison of 0.10 M Be(NO3 )2
solution and 0.10 M CH3COOH solution.
Solution [H3O+] pH
% hydrolysis or
% ionization0.10 M Be(NO3)2 1.0 x 10-3 M 3.00 1.0%
0.10 M CH3COOH 1.3 x 10-3 M 2.89 1.3%
Notice that the Be solution is almost as acidic as the acetic acid solution.
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Synthesis Question Rain water is slightly acidic because it
absorbs carbon dioxide from the atmosphere as it falls from the clouds. (Acid rain is even more acidic because it absorbs acidic anhydride pollutants like NO2 and SO3 as it falls to earth.) If the pH of a stream is 6.5 and all of the acidity comes from CO2, how many CO2 molecules did a drop of rain having a diameter of 6.0 mm absorb in its fall to earth?
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Synthesis Question
Lmol77
32
7
32
77
a
7
32
3a
-332
3222
-75.6
104.2104.2COH
102.4COH
102.3102.3K
102.4COH
HCOHK
HCOHCOH
COH OH CO
102.310H6.5pH
M
M
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Synthesis Question
molecules CO 106.1
mol
molecules106.022mol 106.2
104.2L101.1 molecules CO ofnumber
L101.1cm 11.0
cm 0.3 r droplet water of volume
213
2311
Lmol74
2
41000cm
1L3
33
433
4
3
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Group Question
A common food preservative in citrus flavored drinks is sodium benzoate, the sodium salt of benzoic acid. How does this chemical compound behave in solution so that it preserves the flavor of citrus drinks?
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End of Chapter 18
Weak aqueous acid-base mixtures are called buffers. They are the subject of Chapter 19.