Question

Calculate the pH of a 0.0290 M aqueous solution of the weak base triethylamine ((C2H5)3N, Kb = 5.20×10-4). pH = ?

Answer 1

Equilibrium reaction is -

(C2H5)3N (aq) + H2O(L) -------> (C2H5)3NH+ + OH-

Kb = base dissociation constant

And kb for the above reaction can be written as-

Kb = [(C2H5)3NH+ ] [OH-] / [(C2H5)3N]

Consider [(C2H5)3NH+ = [OH-] = x

So, Kb = x^2/ 0.0290 - x

And if 0.0290 >>>>x then

Kb = x^2/ 0.0290

x = √(Kb × 0.0290)

x = √(5.20×10^-4 × 0.0290) ....(because concentration of solution = 0.0290 M and kb = 5.20×10^-4 are given).

x = 3.88 × 10^-3

Hence X1 = 3.88 × 10^-3

Further if we want to calculate X2 then we can do like this -

X2 = √(5.20×10^-4) × (0.0290 - 3.88×10^-3)

X2 = 3.614×10^-3

Further you can also calculate X3

X3 = √(5.20×10^-4) × (0.0290 -3.614×10^-3)

X3 = 3.633×10^-3

Further

X4 = √(5.20×10^-4) × (0.0290 - 3.633 ×10^-3)

X4 = 3.632 × 10^-3

Now from X1 ,X2, X3 & X4 we can accept

X = 3.63 × 10^-3 mol/litre & = [OH-]

Now, we know pOH = -log [OH-]

pOH = -log(3.63×10^-3)

pOH = 2.44

Further we know pH = 14 - pOH

So,

pH = 14 - 2.44  

pH = 11.56