Question

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

Answer 1

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

3.4×10−4 = x^2 / 1.60-x

Solving for x,

x = [OH-] = 0.023 M

pH = 14 + log 0.023 = 12.36

Therefore, the first option is the closest one.

Answer 2

5.2

Step-by-step 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