Calculating the pH of a salt solution is essential in various scientific fields, including chemistry and environmental science. Salt solutions can affect the acidity or basicity of a solution, depending on their constituent ions. Commonly, salts are formed from the neutralization of an acid and a base, and their pH can be derived from the hydrolysis of these ions in water. In this article, we will explore three diverse examples of calculating the pH of a salt solution. Each example illustrates a different type of salt and its impact on pH.
Sodium acetate (CH₃COONa) is a salt derived from the weak acid acetic acid and the strong base sodium hydroxide. It is commonly used in buffer solutions to maintain a stable pH.
To calculate the pH of a sodium acetate solution, we must consider its hydrolysis in water.
Given: A 0.1 M sodium acetate solution.
The acetate ion hydrolyzes as follows:
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
The equilibrium constant (K_b) for acetate can be calculated using the K_w and K_a of acetic acid:
K_a (acetic acid) = 1.8 x 10⁻⁵, therefore:
K_b = K_w / K_a = 1.0 x 10⁻¹⁴ / 1.8 x 10⁻⁵ = 5.56 x 10⁻¹⁰
Set up the equation for K_b:
K_b = [CH₃COOH][OH⁻] / [CH₃COO⁻]
Let x = [OH⁻], then:
K_b = x² / (0.1 - x) ≈ x² / 0.1
5.56 x 10⁻¹⁰ = x² / 0.1
x² = 5.56 x 10⁻¹¹
x = √(5.56 x 10⁻¹¹) ≈ 7.45 x 10⁻⁶ M
Now, calculate pOH:
pOH = -log(7.45 x 10⁻⁶) ≈ 5.13
Finally, calculate pH:
pH = 14 - pOH = 14 - 5.13 ≈ 8.87
Sodium acetate results in a basic solution due to the hydrolysis of the acetate ion, which produces hydroxide ions.
Ammonium chloride (NH₄Cl) is a salt formed from the weak base ammonia and the strong acid hydrochloric acid. It is often used in fertilizers and as a food additive.
To calculate the pH of an ammonium chloride solution, we analyze the hydrolysis of the ammonium ion.
Given: A 0.2 M ammonium chloride solution.
The ammonium ion hydrolyzes in water:
NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺
The equilibrium constant (K_a) for ammonium can be calculated from K_b of ammonia:
K_b (ammonia) = 1.8 x 10⁻⁵, therefore:
K_a = K_w / K_b = 1.0 x 10⁻¹⁴ / 1.8 x 10⁻⁵ = 5.56 x 10⁻¹⁰
Set up the equation for K_a:
K_a = [NH₃][H₃O⁺] / [NH₄⁺]
Let y = [H₃O⁺], then:
K_a = y² / (0.2 - y) ≈ y² / 0.2
5.56 x 10⁻¹⁰ = y² / 0.2
y² = 5.56 x 10⁻¹¹
y = √(5.56 x 10⁻¹¹) ≈ 7.45 x 10⁻⁶ M
Now, calculate pH:
pH = -log(7.45 x 10⁻⁶) = 5.13
Ammonium chloride results in an acidic solution due to the hydrolysis of the ammonium ion, which produces hydronium ions.
Sodium phosphate (Na₃PO₄) is a salt composed of a strong base (sodium hydroxide) and a weak acid (phosphoric acid). It is commonly used in buffer solutions and as a food additive.
To calculate the pH of a sodium phosphate solution, we need to consider the phosphate ions’ hydrolysis.
Given: A 0.1 M sodium phosphate solution.
The phosphate ion hydrolyzes as follows:
PO₄³⁻ + H₂O ⇌ HPO₄²⁻ + OH⁻
The equilibrium constant (K_b) for phosphate can be calculated using the K_a of dihydrogen phosphate:
K_a (H₂PO₄⁻) = 6.2 x 10⁻⁸, therefore:
K_b = K_w / K_a = 1.0 x 10⁻¹⁴ / 6.2 x 10⁻⁸ = 1.61 x 10⁻⁷
Set up the equation for K_b:
K_b = [HPO₄²⁻][OH⁻] / [PO₄³⁻]
Let z = [OH⁻], then:
K_b = z² / (0.1 - z) ≈ z² / 0.1
1.61 x 10⁻⁷ = z² / 0.1
z² = 1.61 x 10⁻⁸
z = √(1.61 x 10⁻⁸) ≈ 4.01 x 10⁻⁴ M
Now, calculate pOH:
pOH = -log(4.01 x 10⁻⁴) ≈ 3.39
Finally, calculate pH:
pH = 14 - pOH = 14 - 3.39 ≈ 10.61
Sodium phosphate results in a basic solution due to the hydrolysis of the phosphate ion, which produces hydroxide ions.
By exploring these examples of calculating the pH of a salt solution, one can better understand how different salts influence the acidity or basicity of a solution.