Question

5.83 is wrong

Calculate the pH at the equivalence point for the titration of 0.110 M methylamine (CH,NH,) with 0.110 M HCI. The Ky of methylamine is 5.0 x 10- pH = 5.83

Answer 1

Net ionic equation for the titration represented as :-

CH₃NH₂ + H⁺→CH₃NH₃⁺

Divided into two parts :- Stoichiometry problem and equilibrium problem .

Stoichiometry Problem :
At the equivalence point, the number of mole of the acid added is equal to the number of mole of base present.

Since the concentrations of base and acid are equal, the concentration of the conjugate acid CH₃NH₃⁺ can be determined as follows:

Since equal volumes of the acid and base should be mixed, and since they are additive, the concentration of CH₃NH₃⁺ will be half the initial concentration of CH₃NH₂. (means 0.110 M ÷ 2 )

Thus, [CH₃NH₃⁺] = 0.055M

Equilibrium Problem :
The conjugate acid that will be the major species at the equivalence point, will be the only significant source of H⁺ in the solution and therefore, to find the pH of the solution we should find the [H⁺] from the dissociation of CH₃NH₃⁺:

the solution we should find the [H+] from the dissociation of CH3NH+3:

CH₃NH₃⁺ ⇌ CH₃NH₂ + H⁺
Initial: 0.055 M 0 M 0 M
Change: −x M +x M +x M
Equilibrium: (0.055 −x)M x M x M

K_a = [CH₃NH₂] [H⁺] / [CH₃NH₃⁺]

Note that :- K_w = K_a × K_b    ( k_w = 1.0 × 10^-14)

⇒ K_a = K_w K_b = 1.0 × 10^-14 / 5.0×10^-4  = 2.0×10^-11

⇒K_a = [CH₃NH₂] [H⁺]/[CH₃NH₃⁺] = x⋅ x / 0.055 −x

since Ka is small, x will be small compared to 0.075 and we have

= x² / 0.055 = 2.0×10^-11(putting k_a value)

Solve for x = 1.07 ×10^-6 M = [H+]

Therefore, the pH of the solution is

pH= −log[H+]

pH = −log (1.07×10^-6) = 5.971