Calculate the [H⁺] of a 0.50 M solution of methylammonium bromide, CH₃NH₃Br. The K_b of methylamine, CH₃NH₂, is given as 4.4 x 10^-4.

• A. 0.50 M
• B. 1.14 × 10⁻¹¹ M
• C. 3.4 × 10⁻⁶ M
• D. 2.97 × 10⁻⁹ M

Answer: (C)

Verified step by step guidance

1. Identify the chemical species involved: Methylammonium bromide (CH3NH3Br) dissociates in water to form CH3NH3+ and Br-. The CH3NH3+ ion is the conjugate acid of methylamine (CH3NH2).
2. Understand the relationship between Kb and Ka: The Kb value given is for methylamine (CH3NH2). To find the Ka for its conjugate acid (CH3NH3+), use the relation Ka * Kb = Kw, where Kw is the ion-product constant of water (1.0 x 10^-14 at 25°C).
3. Calculate Ka: Rearrange the equation to find Ka = Kw / Kb. Substitute the values: Ka = (1.0 x 10^-14) / (4.4 x 10^-4).
4. Set up the equilibrium expression for CH3NH3+ dissociation: CH3NH3+ ⇌ CH3NH2 + H+. The initial concentration of CH3NH3+ is 0.50 M. Let x be the change in concentration at equilibrium, where [H+] = x.
5. Use the expression for Ka: Ka = [CH3NH2][H+]/[CH3NH3+]. Substitute the equilibrium concentrations: Ka = (x)(x)/(0.50 - x). Assuming x is small compared to 0.50, simplify to Ka ≈ x^2/0.50. Solve for x to find [H+].