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15.11: Détermination du pH des solutions salines
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Determining the pH of Salt Solutions
 
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15.11: Determining the pH of Salt Solutions

The pH of a salt solution is determined by its component anions and cations. Salts that contain pH-neutral anions and the hydronium ion-producing cations form a solution with a pH less than 7. For example, in ammonium nitrate (NH4NO3) solution, NO3 ions do not react with water whereas NH4+ ions produce the hydronium ions resulting in the acidic solution.  In contrast, salts that contain pH-neutral cations and the hydroxide ion-producing anions form a solution with a pH greater than 7. For example, in sodium fluoride (NaF) solution, the Na+ is pH-neutral but the F- produces the hydroxide ions and forms the basic solution. The counterions of a strong acid or base are pH-neutral and salts formed by such counterions form a neutral solution with a pH equal to 7. For example, in KBr, The K+ cation is inert and will not affect pH. The bromide ion is the conjugate base of a strong acid, and so it is of negligible base strength (no appreciable base ionization). The solution is neutral.

Some salts contain both an acidic cation and a basic anion. The overall acidity or basicity of a solution is determined by the relative strength of the cation and anion, which can be compared using Ka and Kb. For example, in NH4F, the NH4+ ion is acidic and the F ion is basic (conjugate base of the weak acid HF). Comparing the two ionization constants: Ka of NH4+ is 5.6 × 10−10 and the Kb of F is 1.6 × 10−11, so the solution is acidic, since Ka > Kb.

Calculating the pH of an Acidic Salt Solution

Aniline is an amine that is used to manufacture dyes. It is isolated as anilinium chloride, [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 anilinium chloride?

Eq1

The Ka for anilinium ion is derived from the Kb for its conjugate base, aniline:

Eq2

Using the provided information, an ICE table for this system is prepared:

C6H5NH3+ (aq) H3O+ (aq) C6H5NH2 (aq)
Initial Concentration (M) 0.233 ~0 0
Change (M) x +x +x
Equilibrium Concentration (M) 0.233 − x x x

Substituting these equilibrium concentration terms into the Ka expression gives

Eq3

Assuming x << 0.233, the equation is simplified and solved for x: 

Eq4

Eq5

The ICE table defines x as the hydronium ion molarity, and so the pH is computed as

Eq6

Hydrolysis of [Al(H2O)6]3+

Calculate the pH of a 0.10 M solution of aluminum chloride, which dissolves completely to give the hydrated aluminum ion [Al(H2O)6]3+ in solution.

The equation for the reaction and Ka are:

Eq7

An ICE table with the provided information is

Al(H2O)63+ (aq) H3O+ (aq) Al(H2O)5(OH)2+ (aq)
Initial Concentration (M) 0.10 ~0 0
Change (M) −x +x +x
Equilibrium Concentration (M) 0.10 − x x x

Substituting the expressions for the equilibrium concentrations into the equation for the ionization constant yields:

Eq8

Assuming x << 0.10 and solving the simplified equation gives:

Eq9

The ICE table defined x as equal to the hydronium ion concentration, and so the pH is calculated to be 2.92, and the solution is acidic.

Eq10

This text is adapted from Openstax, Chemistry 2e, Section 14.4: Hydrolysis of Salts.

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