Question

Calculate the pH of the buffer formed by mixing equal volumes [C₂H₅NH₂]= 1.68M with [HClO₄]= 0.984M . Kb=4.3

Answer 1

Let the volume of each be 1.0 L

Thus, After mixing, the moles of each will be changed, as the volume of the mixture gives 2.0 L

Concentration of [C2H5NH2] before mixing = 1.68 M

Concentration of [C2H5NH2] after mixing = 1.68 mol / 2.0 L = 0.84 M

Concentration of [HClO4] before mixing = 0.984 M

Concentration of [HClO4] after mixing = 0.984 mol / 2.0 L = 0.492 M

Thus, the limiting reagent is [HClO4].

So, from the reaction, C2H5NH2 + HClO4 <----------> C2H5NH3+ + ClO4-

the [C2H5NH3+] produced = 0.492 M

Now, let us find pH of this solution.

C2H5NH3+ + H2O <--------------> C2H5NH2 + H3O+

I 0.492 M 0 0

C -x +x +x

E 0.492 - x +x +x

Ka = [H3O+][C2H5NH2] / [C2H5NH3+]

But, we are given, Kb = 4.3 x 10^-4

Ka = Kw / Kb = (1 x 10^-14) / (4.3 x 10^-4) = 2.33 x 10^-11

2.33 x 10^-11 = x^2 / 0.492 - x

x^2 = 1.144 x 10^-11

x = 3.38 x 10^-6 M

[H3O+] = x = 3.38 x 10^-6 M

pH = -log[H3O+]

pH = -log[3.38 x 10^-6]

pH = 5.47