A weak base ionizes in water as follows: B_(aq) + H₂O_(l) ⇌ BH⁺_(aq)+ OH^-_(aq)
A 0.150 M solution of B is 2.500 % ionized at 25ºC.What is the pH of this solution?

Question

A weak base ionizes in water as follows: B_(aq) + H₂O_(l) ⇌ BH⁺_(aq)+ OH^-_(aq) A 0.150 M solution of B is 2.500 % ionized at 25ºC.What is the pH of this solution?

Quizplus · Accepted Answer

First, we need to find the equilibrium concentration of [H+] in the solution. 
Let x be the concentration of [H+] in mol/L. Then, the concentration of [OH-] is (1.00 x 10^-14)/x.
 
Using the equilibrium expression for the ionization of the weak base, we can set up the following equation:

Ka = ([BH+][OH-])/[B]

where [BH+] = [H+] and [B] = initial concentration of weak base.

Plugging in the given values, we get:

4.46 x 10^-10 = (x^2)/(0.150 - x)

Since x is much smaller than 0.150, we can make the approximation that 0.150 - x ≈ 0.150.

Solving for x, we get:

x = 1.191 x 10^-6 mol/L

Therefore, pH = -log(1.191 x 10^-6) ≈ 2.43.

Thus, the answer is B.

A Weak Base Ionizes in Water as Follows: B(aq)

A Weak Base Ionizes in Water as Follows: B_(aq)