Question

1. Consider these four reactions.

HClO₄ ? H⁺ + ClO₄^-
HCl ? H⁺ + Cl^-
HCN ? H⁺ + CN^-

Determine which of the following is the strongest conjugate base.

a. ClO₄^-

b. Cl^-    

c. CN^-

2. a. Calculate the pH of a 0.6 M solution of HN₃ with K_a = 1.9

Answer 1

1) weaker the acid stronger is the conjugate base.Since the weakest acid among the three is HCN

Therefore, the strongest conjugate base is CN^-

2) HN₃(aq) --------------> N₃^-(aq) + H⁺(aq)

Initially, [HN₃] = 0.6 M & [H⁺] = [N₃^-] = 0 M

Let at eqb. [HN₃] = (0.6 - x) M & [H⁺] = [N₃^-] = x M

Now, K_a = {[H⁺]*[N₃^-]}/[HN₃] = x²/(0.6-x)

or, 1.9*10^-5 = x²/(0.6-x)

or, x² + (1.9*10^-5)x - 1.14*10^-5 = 0

or, x = 3.37*10^-3 M

now, pH = -log[H⁺] = -logx = 2.473

Now, % dissociation = (x/0.6)*100 = [(3.37*10^-3)/0.6]*100 = 0.562 %