Question

The pH of a 0.30 M solution of a weak acid is 2.67. What is the Ka for this acid?
A) 4.4 X10^-4
B) 6.6 x10^-4
C) 1.5 x10^-4
D) 2.1 x 10^-4
E) None of the above

Answer 1

Answer:

Ka  → 1.5×10⁻⁵

Option E. None of the above

Explanation:

We propose the reaction of equlibrium

Weak ac.H +  H₂O ⇄  Weak ac⁻  +  H₃O⁺

Initially we have 0.30 moles of acid in 1 L

In equilibrium we would have:

Weak ac.H +  H₂O ⇄  Weak ac⁻  +  H₃O⁺

0.30 - x                               x               x

We have the pH, where we can obtanined the x, the [H₃O⁺] in the equilibrium.

pH = - log [H₃O⁺] → [H₃O⁺] = 10^⁻(pH)

[H₃O⁺] = 10⁻²'⁶⁷ = 2.14×10⁻³

So let's determine the concentration of the acid, in the equilibrium

0.30 - 2.14×10⁻³ = 0.29786 → [Weak ac.H]

2.14×10⁻³ →  [H₃O⁺] = Conjugate base (Weak ac.⁻)

Let's make the expression for Ka

Ka = [Weak ac.⁻] .  [H₃O⁺]  / [Weak ac.H]

Ka = x² / 0.30 - x

Ka = (2.14×10⁻³)² / 0.29786 → 1.5×10⁻⁵