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Hydrolysis of the type IV salts has one stage regardless of the cation or anion

charge.

weak acid ion
weak base ion
+
–
CH 3COONH 4 ↔ CH 3COO + NH 4
CH 3COONH 4 + H 2O = CH 3COOH + NH 4OH,

+
–
CH 3COO + NH 4 + H 2O = CH 3COOH + NH 4OH.

Hydrolysis of salts of this type occurs completely or almost completely and

is recorded in one stage, regardless of the weak base acidity or the weak acid

basicity. The medium can be considered as close to neutral (pH ≈ 7).

Examples of problem solving
Example 1. 50 g of a solution contains 2.5 g of potassium hydroxide KOH.

Calculate the mass fraction of KOH.
Input: Solution :

m (KOН) = 2,5 g Using the formula for determining the
m solution = 50 g mass fraction

Determine

ω (KOН) – ?

let us determine the KОН mass fraction:

Answer: ω (KOН) = 5 %.

Example 2. Calculate the molar concentration of the sulfate acid solution if

100 ml of this solution contains 4.904 g of H 2SO 4.

Input: Solution :

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