The base dissociation constant or Kb is a value used to measure the strength of a specific base in solution. To determine the percent ionization of the substance we make use of the Kb given. Methylamine or CH3NH2 when in solution would form ions:
CH3NH2 + H2O < = > CH3NH3+ + OH-
Kb is expressed as follows:
Kb = [OH-] [CH3NH3+] / [CH3NH2]
Where the terms represents the concentrations of the acid and the ions.
By the ICE table, we can calculate the equilibrium concentrations,
CH3NH2 CH3NH3+
OH-
I 1.60 0
0
C -x +x
+x
--------------------------------------------------
E 1.60-x x
x
Kb = [OH-] [CH3NH3+] / [CH3NH2] = 3.4×10−4
Solving for x,
x = [OH-] = 0.023 M
pH = 14 + log 0.023 = 12.36
Therefore, the first option is the closest one.
Answer:
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
