Calculate the ph at the equivalence point for the titration of a solution containing 150.0 mg of ethylamine (c2h5nh2) with 0.1000 m hcl solution. the volume of the solution at the equivalence point is 250.0 ml. kb forethylamine is 4.7 × 10−4 .

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.

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.

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.

2H₂O ↔ H₃O⁺ + OH⁻  

Kw = 1.0 x 10⁻¹⁴ = [H₃O⁺][OH⁻]  

[H₃O⁺] = 1.0 x 10⁻⁷ mol/L.

pH = - log [H₃O⁺] = - log (6.32 x 10⁻⁷)

pH = 6.20 .

c, liquids evaporate, creating a gas