L8 - Salt Hydrolysis

objectives

• Prediction of the relative pH of aqueous salt solutions formed by the different combinations of strong and weak acid and base.

Recap of Strong Acids and Bases

Remember Strong Acids and Bases Completely dissociate in solution.

This means the ions do not like bonding to each other and therefore will remain as spectator ions in water rather than reform the acid or base.

For the IB we are only expected to know:

• Strong Bases are All Group 1 Hydroxides and Barium Hydroxide - All other Hydroxides (such as Calcium Hydroxide) are classed as weak bases. The most common Bases in the IB are Amines and Ammonia.
• Strong Acids are limited to Halogen Halides (i.e. HCl, HI, HBr, HF), HNO3, H2SO4 and H3PO4. All other acids are weak acids - these include Carbonic Acid (H2CO3) and Carboxylic Acids.

Salt Hydrolysis

'a reaction in which the cation or anion or both of a salt react with water to produce acidity or alkalinity'

• Salts of weak acids and weak bases can react with water
• The result of these reactions is a solution which is not neutral

Example 1: Iron (II) Chloride FeCl₂

In this example the Chloride ion is a spectator ion and cannot convert to HCl. This is because the Cl- is from a strong acid and as strong acids completely dissociate it cannot reform HCl.

Fe⁺² + 2H₂O ⇋ Fe(OH)₂ + 2H⁺

As the Reaction is in equilibrium it is only partially converted to H+. This means that the solution would become slightly acidic.

Example 2: Sodium Ethanoate CH₃COONa

In this example the Sodium behaves as a spectator ion as it is from a Strong Base and will not convert to NaOH.

CH₃COO^- + H₂O ⇋ CH₃COOH + OH^-

As the reaction is in equilibrium it is only partially converted to OH-. This means that the solution would become slightly alkaline.

Predicting the pH of Salt Solutions

Just like with the shapes of pH curves - we can predict the pH of salt solutions if we know their constituent acid and base.

1. The Salt of a Strong Acid and a Weak Base

Examples: FeCl₃, CuCl₂, AlCl₃, NH₄Cl, CuSO₄

• The solution of such a salt is acidic in nature.
• The cation of the salt which has come from weak base is reactive.
• It reacts with water to form a weak base and H+ ions.

B⁺ + H₂O ↔ BOH + H⁺

• Consider, for example, NH₄Cl. It ionises in water completely into NH₄⁺ and Cl ions. ions react with water to form a weak base (NH₄OH) and H⁺ ions.

NH₄⁺ + H₂O ↔ NH₄OH + H⁺

Proving this Mathematically

We can use the fundamental relationship:

• Ka x Kb = Kw

We must remember that the easiest way to solve these questions is to follow the following steps:

Strong Acid and Weak Base

1. Write an Equation for the Salt reacting with Water - remember in this example the salt is a weak acid.

2. Rearrange for Ka

Ka = Kw/Kb

3. Substitute numbers and find Ka

4. Use our Weak acid expression to calculate [H⁺]

[H⁺] = √Ka x [Salt]

5. Use the pH Expression to find pH

pH = -log₁₀[H⁺]

Example Question:

Aniline is an amine that is used to manufacture dyes.

It is isolated as aniline hydrochloride, [C₆H₅NH₃⁺]Cl, a salt prepared by the reaction of the weak base aniline and hydrochloric acid.

What is the pH of a 0.233 M solution of aniline hydrochloride given Kb for C₆H₅NH₂ = 4.3 x10^-10?

Step 1: Write the equation for the reaction of [C₆H₅NH₃⁺]Cl with Water

C₆H₅NH₃⁺ + H₂O ⇋ C₆H₅NH₂ + H₃O⁺

Step 2 & 3: as Kb for C₆H₅NH₂ = 4.3 x10^-10

Rearrange for Ka and substitute Kw and Kb

Ka = Kw/Kb

Ka = 1x10^-14/4.3x10^-10

Ka = 2.3x10^-5

Step 4: Use the expression for Ka and rearrange for H+

[H⁺] = √Ka x [Salt]

[H⁺] = √2.3x10^-5 x 0.233

[H⁺] = 2.3x10^-3

Step 5: Calculate pH

pH = -log₁₀[H⁺]

pH = -log₁₀[2.3x10^-3]

pH = 2.64

2. The Salt of a Strong Base and a Weak Acid

Examples: CH₃COONa, NaCN, NaHCO₃, Na₂CO₃

• The solution of such a salt is basic in nature.
• The anion of the salt which has come from weak acid is reactive.
• It reacts with water to form a weak acid and OH- ions.

A^- + H₂O ⇋ AH + OH^-

• Consider, for example, CH₃COONa. It ionises in water completely into CH₃COO^- and Na ions. ions react with water to form a weak acid (CH₃COOH) and OH^- ions.

CH₃COO^- + H₂O ⇋ CH₃COOH + OH^-

Proving this Mathematically

We can use the fundamental relationship:

• Ka x Kb = Kw

We must remember that the easiest way to solve these questions is to follow the following steps:

Weak Acid and Strong Base

1. Write an Equation for the Salt reacting with Water - remember in this example the salt is a weak base.

2. In this example you can either convert Ka to Kb to get Kb or directly use Kb

Kb = Kw/Ka

3. Substitute numbers and find Kb

4. Use our Weak base expression to calculate [OH^-]

[OH^-] = √Kb x [Salt]

5. Use the pH = 14 - pOH expression to find pH

pH = 14 - pOH

Example Question

What is the pH of a 0.083-M solution of CN−?

Use 4.9 × 10−10 as Ka for HCN.

1. Write an Equation for the Salt reacting with Water - remember in this example the salt is a weak base.

CN^- + H₂O ⇋ HCN + OH^-

2. In this example you can either convert Ka to Kb to get Kb

Kb = Kw/Ka

3. Substitute numbers and find Kb

Kb = 1x10^-14/ 4.9x10^-10

Kb = 2x10^-5

4. Use our Weak base expression to calculate [OH^-]

[OH^-] = √Kb x [Salt]

[OH^-] = √(2x10^-5 x 0.083)

[OH^-] = 1.3x10^-3

5. Use the pH = 14 - pOH expression to find pH

pH = 14 - pOH

pH = 14 - 2.89

pH = 11.11

3. Salts of Strong Acids and Strong Bases

A strong acid and a strong base, such as HCl(aq) and NaOH(aq) will react to form a neutral solution since the conjugate partners produced are of negligible strength.

Na and Cl ions will not react with water and so this means the pH will remain at 7.

Examples in the DP are Sodium and Potassium Salts of HCl, H₂SO₄, HNO₃ & H₃PO₄

4. Salts of Weak Acids and Weak Bases

As the salts of these neutralisations will make a weak base and a weak acid we will have to compare the relative strength of weak acids and bases using Ka and Kb.

• Where Ka > Kb the Solution will be acidic (pH < 7)
• Where Kb > Ka the Solution will be alkaline (pH > 7)
• Where Ka = Kb the Solution will be Neutral (pH = 7)

What about Amphiprotic Species?

As these substances can both accept and donate protons we can compare their relative acid and base strengths using the Ka and Kb for the same species.

Example: HCO₃^-

has a Ka of 4.7x10^-11

This means (using Kb = Kw/Ka) it has a Kb of 2.3x10^-8

Therefore Kb > Ka

So the pH will be alkaline