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
