Calculate the ph of a 1.60 mch3nh3cl solution. kb for methylamine, ch3nh2, is 3.4×10−4m. calculate the of a 1.60 solution. for methylamine, , is . 12.39 8.82 1.61 5.18

5.2

Explanation:

Calculate the pH of a 1.60 M CH₃NH₃Cl solution. Kb for methylamine, CH₃NH₂, is 3.7 × 10⁻⁴.

CH₃NH₃Cl is a strong electrolyte that ionizes according to the following equation.

CH₃NH₃Cl(aq) → CH₃NH₃⁺(aq) + Cl⁻(aq)

The concentration of CH₃NH₃⁺ will also be 1.60 M (Ca). CH₃NH₃⁺ is the conjugate acid of CH₃NH₂. We can find its acid dissociation constant (Ka) using the following expression.

Ka × Kb = Kw

Ka × 3.7 × 10⁻⁴ = 1.0 × 10⁻¹⁴

Ka = 2.7 × 10⁻¹¹

The acid dissociation of CH₃NH₃⁺ can be represented through the following equation.

CH₃NH₃⁺(aq) ⇄ CH₃NH₂(aq) + H⁺(aq)

For a weak acid, we can find the concentration of H⁺ using the following expression.

[H⁺] = √(Ka × Ca) = √(2.7 × 10⁻¹¹ × 1.60) = 6.6 × 10⁻⁶ M

The pH is:

pH = -log [H⁺] = -log (6.6 × 10⁻⁶) = 5.2