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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 acidKa=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)

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?

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?

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.

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