Question

Part A Calculate the molar concentration of OH− ions in a 8.1×10−2 M solution of ethylamine (C2H5NH2)(Kb=6.4×10−4). Express your answer using two significant figures. [OH−] = M

Part B Calculate the pH of this solution. Express your answer using two decimal places.

Answer 1

Ans. Balanced reaction:        C₂H₅NH₂(aq) + H₂O(l) <----------> C₂H₅NH₃⁺(aq) + OH^-(aq)

Given, Initial [C₂H₅NH₂] = 8.1 x 10^-2 M = 0.081 M

            Kb of ethylamine (C₂H₅NH₂) = 6.4 x 10^-4

# Create an ICE table with initial [C₂H₅NH₂] as shown in figure-

Now,

            Base dissociation constant, Kb = [C₂H₅NH₃⁺] [OH^-] / [C₂H₅NH₂]

            Or, 6.4 x 10^-4 = (X) (X) / (0.081 - X)²

                Or, 6.4 x 10^-4 (0.006561+ X² – 0.162X) = X²

            Or, 4.19904 x 10^-6 + (6.4 x 10^-4)X² – (1.0368 x 10^-4)X = X²

            Or, X² - 4.19904 x 10^-6 - (6.4 x 10^-4)X² + (1.0368 x 10^-4)X = 0

            Or, 0.99936X² + (1.0368 x 10^-4)X - 4.19904 x 10^-6 = 0

Solving the quadratic equation we get following roots-

            X1 = 0.001998                      ; X2 = -0.02055

Since concentration can’t be negative, reject X2.

Hence, X = 0.001998

#1. Now, in the ICE, [OH^-] at equilibrium = X = 0.001998

Hence, equilibrium [OH^-] = 0.001998 M = 1.998 x 10^-3 M = 2.0 x 10^-3 M

#2. pOH = -log [OH^-] = - log (0.001998) = 2.699

Using,             pH + pOH = 14.0

            Or, pH = 14.00 – pOH = 14.00 – 2.699 = 11.301

Hence, pH of 0.081 M ethylamine = 11.301