AP*
Chapter 14
Acids and Bases
Section 14.8
Acid-Base Properties of Salts
Salts
 Ionic compounds.
 When dissolved in water, break up into its ions (which
can behave as acids or bases).
 Review the solubility rules (p. 156, table 4.1)
 Salts containing NO3-, NH4+ and group 1A cations are
highly soluble.
 Example:
KCl (s) + H2O (l)  K+ + Cl- + H2O (l)
Net equation: KCl (s)  K+ + Cl
Note, this reaction goes to completion, due to the high solubility of KCl.
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Section 14.8
Acid-Base Properties of Salts
Salts
 The dissolved ions of salts can affect pH.
 The salt of a strong acid and a strong base gives a
neutral solution. Only starting ions will be present in significant
quantities, so pH will be neutral (due to H2O only)
 For example, KCl & NaNO3 will NOT change the pH.
 Why is this true?
KCl (s)  K+ + ClK+ + H2O   KOH + H+
Cl- + H2O   HCl + OHSince K+, Cl- , Na+ & NO3- are all conjugates of strong
acids/bases, these ions will not change the pH.
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Section 14.8
Acid-Base Properties of Salts
Salts
 A basic solution is formed if the anion of the salt is the
conjugate base of a weak acid.
 For example, NaF & KC2H3O2 WILL change the pH.
 Why is this true?
NaF (s)  Na+ + FNa+ + H2O   NaOH + H+
F- + H2O   HF + OHSince Na+ is a conjugate of strong base = neutral pH
Since F- is a conjugate of weak acid (HF), pH will be >7,
because Na+, F-, H2O, HF & OH- will all be present in solution.
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Section 14.8
Acid-Base Properties of Salts
Salts
 An acidic solution is formed if the cation of the salt is
the conjugate acid of a weak base.
 NH4Cl will produce an acidic solution.
 Why is this true?
NH4Cl (s)  NH4+ + ClNH4+ + H2O   NH3 + H3O+
Cl- + H2O   HCl + OHSince NH4+ is a conjugate of weak base, pH will be <7,
because NH4+, H2O, NH4Cl & H3O+ will all be present
in solution.
Since Cl- is a conjugate of strong acid = neutral pH
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Section 14.8
Acid-Base Properties of Salts
Cation
neutral
neutral
Anion
neutral
conjugate
base of
weak acid
neutral
Acidic
or Basic
neutral
basic
conjugate
acidic
acid of
weak base
conjugate conjugate depends
acid of
base of
on Ka & Kb
weak base weak acid
values
*
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Example
NaCl
NaF
NH4Cl
Al2(SO4)3
6
Section 14.8
Acid-Base Properties of Salts
Qualitative Prediction of pH of Salt Solutions (from Weak
Parents)
If a salt contains
BOTH the
conjugate of a
weak acid & the
conjugate of a
weak base, then
the Ka & Kb
values must be
compared to
determine the
approximate pH.
*
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Section 14.8
Acid-Base Properties of Salts
EXERCISE!
HC2H3O2
Ka = 1.8 × 10-5
HCN
Ka = 6.2 × 10-10
Calculate the Kb values for: C2H3O2− and CN−
Don’t forget: Ka × Kb = Kw
Kb (C2H3O2-) = 5.6 × 10-10
Kb (CN-) = 1.6 × 10-5
Note that as Ka increases, a conjugate Kb decreases!!
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Section 14.8
Acid-Base Properties of Salts
CONCEPT CHECK!
Arrange the following 1.0 M solutions from lowest to highest pH.
HBr
NaCN
NaCl
NaOH
NH3
HF
NH4Cl
HCN
Justify your answer.
HBr, HF, HCN, NH4Cl, NaCl, NaCN, NH3, NaOH
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Section 14.8
Acid-Base Properties of Salts
CONCEPT CHECK!
Consider a 0.30 M solution of NaF.
The Ka for HF is 7.2 × 10-4.
What are the major species?
Na+, F-, H2O
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Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
 Q: Why isn’t NaF considered a major species?
 Answer: Because NaF is highly soluble. It will dissolve
completely into its ions.
 What are the possibilities for the dominant reactions?
 Answer: examine the next slide...
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Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
The possibilities for the dominant reactions are:
1.
2.
3.
4.
F–(aq) + H2O(l)
H2O(l) + H2O(l)
Na+(aq) + H2O(l)
Na+(aq) + F–(aq)
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HF(aq) + OH–(aq)
H3O+(aq) + OH–(aq)
NaOH + H+(aq)
NaF
12
Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
 How do we decide which reaction controls the pH?
 Answer: only reactions 1 & 2 can control pH. Reaction 3
produces a srong base (unreasonable) & reaction 4 will
not be in equilibrium, since all the NaF will dissolve.
 This leaves:
F–(aq) + H2O(l)
HF(aq) + OH–(aq)
H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
 Determine the equilibrium constant for each reaction.
 The first reaction will dominate, due to the large Ka.
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Section 14.8
Acid-Base Properties of Salts
EXERCISE!
Calculate the pH of a 0.75 M aqueous solution of NaCN.
Ka for HCN is 6.2 × 10–10.
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14
Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
 What are the major species in solution?
Na+, CN–, H2O
 Why isn’t NaCN considered a major species?
Because it is highly soluble & will dissolve completely.
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Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
 What are all possibilities for the dominant reaction?
 The possibilities for the dominant reaction are:
1. CN–(aq) + H2O(l)
HCN(aq) + OH–(aq)
2. H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
3. Na+(aq) + H2O(l)
NaOH + H+(aq)
4. Na+(aq) + CN–(aq)
NaCN
 Which of these reactions really occur?
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Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
 How do we decide which reaction controls the pH?
CN–(aq) + H2O(l)
H2O(l) + H2O(l)
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HCN(aq) + OH–(aq)
H3O+(aq) + OH–(aq)
17
Section 14.8
Acid-Base Properties of Salts
Steps Toward Solving for pH
CN–(aq) + H2O
Initial
Change
Equilibrium
HCN(aq) + OH–(aq)
0.75 M
0
~0
–x
+x
+x
0.75–x
x
x
Kb = 1.6 × 10–5
pH = 11.54