Question

Calculate the pH at 25°C of a 0.76M solution of anilinium chloride C6H5NH3Cl. Note that aniline C6H5NH2 is a weak base with a pKb of 4.87.

Answer 1

Answer

pH = 4.63

Explanation

Conjucate acid is partly dissociates in water

C6H5NH3⁺(aq) + H2O(l) -------> C6H5NH2(aq) + H3O⁺(aq)

Ka = [C6H5NH2][H3O⁺]/[C6H5NH3⁺]

Ka = Kw/Kb = 1.00 ×10^-14/1.35×10^-5= 7.41× 10^-10

at equillibrium

[C6H5NH3⁺] = 0.76 - x

[C6H5NH2] = x

[H3O⁺] = x

so,

x²/(0.76 - x) = 7.41×10^-10

we can assume , 0.76 -x = 0.76 because x is small value

x²/0.76 = 7.41 ×10^-10

x² = 5.63× 10^-10

x = 2.37×10^-5

[H3O⁺] = 2.37×10^-5M

pH = -log[H3O⁺]

pH = -log(2.37 ×10^-5)

pH = 4.63