Answer:
12
Explanation:
Let's consider the basic reaction of methylamine.
CH₃NH₂ + H₂O ⇄ CH₃NH₃⁺ + OH⁻
The concentration of the weak base is Cb = 0.195 M and the basic dissociation constant is Kb = 4.4 × 10⁻⁴. We can calculate the concentration of OH⁻ using the following expression.
[OH⁻] = √(Kb × Cb)
[OH⁻] = √(4.4 × 10⁻⁴ × 0.195) = 9.3 × 10⁻³ M
The pOH is:
pOH = -log [OH⁻] = -log 9.3 × 10⁻³ = 2.0
The pH is:
pH + pOH = 14
pH = 14 - pOH = 14 - 2.0 = 12
The pH of the 0.195 M methylamine solution has been 12.0.
The pH of the solution has been defined as the hydrogen ion concentration in the sample. The pOH has been the hydroxide ion concentration in the sample.
From the Kb of the methylamine, the hydroxide ion concentration of the solution can be determined as:
[[tex]\rm OH^-[/tex]] = [tex]\rm \sqrt{Kb\;\times\;concentration}[/tex]
[[tex]\rm OH^-[/tex]] = [tex]\rm \sqrt{4.4\;\times\;10^-^4\;\times\;0.195}[/tex]
[[tex]\rm OH^-[/tex]] = 9.3 [tex]\rm \times\;10^-^3[/tex] M
The pOH of the solution can be calculated as:
pOH = -log [[tex]\rm OH^-[/tex]]
pOH = -log 9.3 [tex]\rm \times\;10^-^3[/tex]
pOH = 2.0
The sum of pH and pOH has been a constant quantity.
pH + pOH = 14
pH + 2 = 14
pH = 14 - 2
pH = 12.0
The pH of the 0.195 M methylamine solution has been 12.0.
For more information about the pH of the solution, refer to the link:
https://brainly.com/question/20437978
