Answer:
pH = 6.2.
Explanation:
• Firstly, we need to calculate the no. of moles (n) of ethylamine (C₂H₅NH₂) using the law:
n = mass / molar mass,
Mass of C₂H₅NH₂ = 150.0 mg = 0.150 g.
Molar mass of C₂H₅NH₂ = 45.0847 g/mol.
∴ The no. of moles (n) of C₂H₅NH₂ = mass / molar mass = (0.150 g) / (45.0847 g/mol) = 0.00333 mol.
- The ionic equation of the equivalence of ethylamine (C₂H₅NH₂) with HCl:
CH₃CH₂NH₂ + H⁺ ↔ CH₃CH₂NH₃⁺
The no of moles of CH₃CH₂NH₃⁺ = 0.00333 mol.
The molarity of (CH₃CH₂NH₃⁺) can be calculated by dividing its no. of moles (0.00333 mol) by the volume of the solution (0.250 L).
[CH₃CH₂NH₃⁺] = 0.00333 mol/ 0.250 L = 0.0133 M.
- There is an equilibrium between the resulting (CH₃CH₂NH₃⁺) and water:
CH₃CH₂NH₃⁺ + H₂O ↔ CH₃CH₂NH₂ + H₃O⁺.
The CH₃CH₂NH₃⁺ decomposed by an amount x and (CH₃CH₂NH₂ & H₃O⁺) formed by amount x.
The hydrolysis constant Ka = Kw / Kb = (1.0 x 10⁻¹⁴) / (4.7 x 10⁻⁴) = 2.1 x 10⁻¹¹.
At equilibrium:
[CH₃CH₂NH₃⁺] = 0.0133 - x
[H₃O⁺] = [CH₃CH₂NH₂] = x
Ka = [CH₃CH₂NH₂] [H₃O⁺] / [CH₃CH₂NH₃⁺] = (x)(x) / (0.0133 -x)
2.1 x 10⁻¹¹ = (x)(x)/ 0.0133-x
x = [H₃O⁺] = 5.32 x 10⁻⁷ mol/L.
- Also, we cannot neglect the [H₃O⁺] from the water dissociation
2H₂O ↔ H₃O⁺ + OH⁻
Kw = 1.0 x 10⁻¹⁴ = [H₃O⁺][OH⁻]
[H₃O⁺] = 1.0 x 10⁻⁷ mol/L.
- The total concentration of (H₃O⁺) = 5.32 x 10⁻⁷ + 1.0 x 10⁻⁷ = 6.32 x 10⁻⁷ mol/L.
pH = - log [H₃O⁺] = - log (6.32 x 10⁻⁷)
pH = 6.20
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