The ammonium ion NH₄⁺ acts as a weak acid according to the equation NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) The [H₃O⁺] of a 0.200 M ammonium chloride solution is closest to A - VCE - SSCE Chemistry - Question 21 - 2016 - Paper 1

To find the hydronium ion concentration, we start from the equilibrium expression for the weak acid dissociation:

1. Identify the dissociation reaction: The reaction is given by: $NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)$

2. Establish equilibrium concentrations: For a 0.200 M solution of ammonium chloride, at equilibrium, let the change in concentration of H₃O⁺ and NH₃ be 'x'. Therefore:

   • [NH₄⁺] at equilibrium = 0.200 - x
   • [NH₃] at equilibrium = x
   • [H₃O⁺] at equilibrium = x

3. Write the expression for the acid dissociation constant, Ka: $K_a = \frac{[NH₃][H₃O⁺]}{[NH₄⁺]}$

   Using the value of Ka for NH₄⁺, which is approximately 5.55 x 10⁻¹⁰: $K_a = \frac{x imes x}{0.200 - x}$ Assuming x is small compared to 0.200, we can simplify to: $K_a = \frac{x^2}{0.200}$

4. Solve for x: Substituting the value of Ka: $5.55 x 10^{-10} = \frac{x^2}{0.200}$ Rearranging gives: $x^2 = 5.55 x 10^{-10} \times 0.200$ $x^2 = 1.11 x 10^{-10}$ $x = \sqrt{1.11 x 10^{-10}} = 1.06 x 10^{-5} M$

5. Conclusion: Thus, the concentration of H₃O⁺ in the solution is approximately 1.06 x 10⁻⁵ M, which corresponds to option C.