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

Assuming minor dissociation due to the weak acidity, let x be the concentration of H₃O⁺ produced:

• Initial [NH₄⁺] = 0.200 M
• Change: [NH₄⁺] = 0.200 - x, [H₃O⁺] = x, and [NH₃] = x.
• At equilibrium:

$K_a = \frac{x^2}{0.200 - x}$

For NH₄⁺, we can use a common Kₐ value around 5.56 x 10⁻¹⁰. Rearranging gives:

$x^2 = K_a (0.200 - x)$ Assuming x is small, we approximate: $x^2 = K_a \cdot 0.200$

Calculate:

$x^2 = (5.56 \times 10^{-10})(0.200) = 1.112 \times 10^{-10}$

Solving for x:

$x = \sqrt{1.112 \times 10^{-10}} \approx 1.06 \times 10^{-5} M$

Thus, the closest value to [H₃O⁺] = 1.06 x 10⁻⁵ M is:

C. 1.06 x 10⁻⁵ M