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Hydrolysis - acidic, basic, and neutral salts
Skills to develop
- Predict the acidity of a salt solution.
- Calculate the pH of a salt solution.
- Calculate the concentrations of various ions in a salt solution.
- Explain hydrolysis reactions.
Acidic, Basic, and Neutral Salts
Ions of Neutral Salts
| Cations
|
|---|
| Na+ | K+ | Rb+ | Cs+
|
| Mg2+ | Ca2+ | Sr2+ | Ba2+
|
| Anions
|
|---|
| Cl- | Br- | I-,
|
| ClO4- | BrO4- | ClO3-
| NO3-
|
A salt is formed between the reaction of an acid and a base.
Usually, a neutral salt is formed when a strong acid and a strong base
is neutralized in the reaction:
H+ + OH- = H2O
The bystander ions in an acid-base reaction form a salt solution.
Most neutral salts consist of cations and anions listed
in the table on the right. These ions have little tendency to react
with water. Thus, salts consisting of these ions are neutral salts.
For example: NaCl, KNO3, CaBr2, CsClO4
are neutral salts.
| Acidic Ions
|
|---|
| NH4+ | Al3+ | Pb2+ | Sn2+
|
| Transition metal ions
|
| HSO4- | H2PO4-
|
| Basic Ions
|
|---|
| F- | C2H3O2-
| NO2- | HCO3-
|
| CN- | CO32-
| S2- | SO42-
|
| HPO42- | PO43-
|
When weak acids and bases react, the relative strength of the conjugated
acid-base pair in the salt determines the pH of its solutions.
The salt, or its solution, so formed can be acidic, neutral or basic.
A salt formed between a strong acid and a weak base is an acid salt,
for example NH4Cl. A salt formed between a weak acid and a strong base is a
basic salt, for example NaCH3COO. These salts are acidic
or basic due to their acidic or basic ions as shown in the tables here.
Hydrolysis of Acidic Salts
A salt formed between a strong acid and a weak base is an acid salt.
Ammonia is a weak base, and its salt with any strong acid gives a
solution with a pH lower than 7. For example, let us consider the
reaction:
HCl + NH4OH = NH4+ + Cl- + H2O
In the solution, the NH4+ ion reacts with water
(called hydrolysis) according to the equation:
NH4+ + H2O = NH3 + H3O+.
The acidity constant can be derived from Kw and
Kb.
[H3O+] [NH3] [OH-]
Ka = ---------------- ------
[NH4+] [OH-]
= Kw / Kb
= 1.00e-14 / 1.75e-5 = 5.7e-10.
Example 1
What is the concentration of NH4+, NH3, and H+ in a
0.100 M NH4NO3 solution?
Solution
Assume that [NH3] = x, then [H3O+] = x, and
you write the concentration below the formula in the reaction:
NH4+ + H2O = NH3 + H3O+
0.100-x x x
Ka = 5.7E-10.
x2
= -------
0.100-x
Since the concentration has a value much greater than Ka, you may use
x = (0.100*5.7E(-10))1/2
= 7.5E-6
[NH3] = [H+] = x = 7.5E-6 M
pH = -log7.5e-6 = 5.12
[NH4+] = 0.100 M
Discussion
Since pH = 5.12, the contribution of [H+] due to self ionization
of water may therefore be neglected.
Hydrolysis and Basic Salts
A basic salt is formed between a weak acid and a strong base.
The basicity is due to the hydrolysis of the conjugate base of the (weak)
acid used in the neutralization reaction.
For example, sodium acetate formed between the weak acetic acid
and the strong base NaOH is a basic salt. When the salt is dissolved,
ionization takes place:
NaAc = Na+ + Ac-
In the presence of water, Ac- undergo hydrolysis:
H2O + Ac- = HAc + OH-
And the equilibrium constant for this reaction is Kb
of the conjugate base Ac- of the acid HAc. Note the following
equilibrium constants:
| N o t e
|
|---|
| Acetic acid | Ka=1.75e-5
|
| Ammonia | Kb=1.75e-5
|
[HAc] [OH-]
Kb = -----------
[Ac-]
[HAc] [OH-] [H+]
Kb = ----------- ---
[Ac-] [H+]
[HAc] [OH-][H+]
Kb = ---------- ---------
[Ac-] [H+]
= Kw / Ka
= 1.00e-14 / 1.75e-5 = 5.7e-10.
Thus,
Ka Kb = Kw
or
pKa + pKb = 14
for a conjugate acid-base pair. Let us look at a numerical problem of this
type.
Example 2
Calculate the [Na+], [Ac-], [H+] and [OH-] of a solution
of 0.100 M NaAc (at 298 K). (Ka = 1.8E-5)
Solution
Let x represent [H+], then
H2O + Ac- = HAc + OH-
0.100-x x x
x2
--------- = (1E-14)/(1.8E-5) = 5.6E-10
0.100-x
Solving for x results in
x = sqrt( 0.100*5.6E-10)
= 7.5E-6
[OH-] = [HAc] = 7.5E-6
[Na+] = 0.100 F
Discussion
This corresponds to a pH of 8.9 or [H+] = 1.3E-9.
Note that Kw / Ka = Kb of Ac-, so that Kb rather
than Ka may be given as data in this question.
Salts of weak acids and weak bases
A salt formed between a weak acid and a weak base can be neutral, acidic,
or basic depending on the relative strengths of the acid and base.
If Ka(cation) > Kb(anion) the solution of the salt is acidic.
If Ka(cation) = Kb(anion) the solution of the salt is neutral.
If Ka(cation) < Kb(anion) the solution of the salt is basic.
Example 3
Arrange the three salts according to their acidity.
NH4CH3COO (ammonium acetate),
NH4CN (ammonium cyanide),
and NH4HC2O4 (ammonium oxalate).
Ka(acetic acid) = 1.85E-5,
Ka(hydrogen cyanide) = 6.2E-10,
Ka(oxalic acid) = 5.6E-2,
Kb(NH3) = 1.8E-5.
Solution
ammonium oxalate -- acidic, Ka(o) > Kb(NH3)
ammonium acetate -- neutral Ka = Kb
ammonium cyanide -- basic Ka(c) < Kb(NH3)
Confidence Building Questions
-
The reaction of an acid and a base always produces a salt
as the by-product, true or false? (t/f)
Answer true
Consider...
Water is the real product, while the salt is formed from the
spectator ions.
-
Is a solution of sodium acetate acidic, neutral or basic?
Answer basic
Consider...
Acetic acid is a weak acid that forms a salt with a strong base, NaOH.
The salt solution turns bromothymol-blue blue.
-
Are solutions of ammonium chloride acidic, basic or neutral?
Answer acidic
Consider...
Ammonium hydroxide does not have the same strength as a base as HCl
has as an acid.
Amonium chloride solutions turn bromothymol-blue yellow.
-
Calculate the pH of a 0.100 M KCN solution.
Ka(HCN) = 6.2e-10, Kb(CN-) = 1.6E-5.
Answer 11.1
Consider...
KCN = K+ + CN-
CN- + H2O = HCN + OH-
(0.100-x) x x
x = (0.100*1.5E-5)1/2
= 1.2E-3
pOH = 2.9
pH = 11.1
-
The symbol Kb(HS-) is the equilibrium constant for the reaction:
a. HS- + OH- = S2- + H2O
b. HS- + H2O = H2S + OH-
c. HS- + H2O = H3O+ + S2-
d. HS- + H3O+ = H2S + H2O
Answer b
Consider...
Write an equation for Kb yourself. Do not guess.
The b. is the closest among the four.
-
What symbol would you use for the equilibrium constant of
HS- = H+ + S2-
Answer Ka
Consider...
This is the ionization of HS-;
Ka for HS-, or
Ka2 for H2S.
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