What is the hydronium-ion concentration at equilibrium in a 0.10 M solution of aniline (C₆H₅NH₂,K_b = 4.2 × 10^-10)at 25^oC? A)1.0 × 10^-13 M B)1.0 × 10^-1 M C)1.5 × 10^-9 M D)6.5 × 10^-6 M E)6.5 × 10^-5 M

Question

Quizplus · Accepted Answer

Aniline is a weak base, so it will react with water to form its conjugate acid and hydroxide ions. The balanced chemical equation for this reaction is: C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-. 
The equilibrium expression for this reaction is: Kb = [C6H5NH3+][OH-]/[C6H5NH2]. 
We are given that [C6H5NH2] = 0.10 M, and we can assume that [OH-] is negligible compared to [C6H5NH3+], so we can simplify the equilibrium expression to: Kb = [C6H5NH3+]/[C6H5NH2]. 
We can then solve for [C6H5NH3+]: [C6H5NH3+] = Kb[C6H5NH2]. 
The value of Kb for aniline is given as 4.2 × 10^-10. Plugging in the values: [C6H5NH3+] = (4.2 × 10^-10)(0.10) = 4.2 × 10^-11 M. 
But remember, this is the concentration of the hydronium ion, not the conjugate acid. So the hydronium-ion concentration is 4.2 × 10^-11 M, which is closest to answer choice C, 1.5 × 10^-9 M.

What Is the Hydronium-Ion Concentration at Equilibrium in a 0