The [H₃O⁺] of a 0.49 M solution of NH₄Cl in H₂O at 25°C is (K_b for NH₃ = 1.8 $\times$ 10^-⁵): A)2.7 $\times$ 10^-¹⁰M B)3.0 $\times$ 10^-³M C)1.6 $\times$10^-⁵M D)0.49 M E)none of these

Question

Quizplus · Accepted Answer

The correct answer is C because the [H₃O⁺] of a 0.49 M solution of NH₄Cl in H₂O at 25°C can be calculated using the Kb expression for NH₃ and the fact that NH₄Cl is a weak acid that undergoes the following equilibrium reaction:NH₄Cl(aq) + H₂O(l) <=> NH₄⁺(aq) + Cl^-(aq)The Kb expression for NH₃ is:Kb = [NH₃][H₃O⁺]/[NH₄⁺]Since NH₄Cl is a weak acid, the [H₃O⁺] will be small compared to the [NH₄⁺]. Therefore, we can approximate the [NH₄⁺] to be equal to the initial concentration of NH₄Cl, which is 0.49 M.Substituting the values into the Kb expression, we get:1.8 x 10^-5 = [NH₃][H₃O⁺]/0.49Solving for [H₃O⁺], we get:[H₃O⁺] = 1.6 x 10^-5 M

The [H3O+] of a 0 ×\times× 10-5): A)2 ×\times× 10-10 M B)3 ×\times×

The [H₃O⁺] of a 0 $\times$ 10^-⁵):
A)2 $\times$ 10^-¹⁰M
B)3 $\times$