Answer:
3.4 mg of methylamine
Explanation:
To do this, we need to write the overall reaction of the methylamine in solution. This is because all aqueous solution has a pH, and this means that the solutions can be dissociated into it's respective ions. For the case of the methylamine:
CH₃NH₂ + H₂O <-------> CH₃NH₃⁺ + OH⁻ Kb = 3.7x10⁻⁴
Now, we want to know how many grams of methylamine we have in 100 mL of this solution. This is actually pretty easy to solve, we just need to write an ICE chart, and from there, calculate the initial concentration of the methylamine. Then, we can calculate the moles and finally the mass.
First, let's write the ICE chart.
CH₃NH₂ + H₂O <-------> CH₃NH₃⁺ + OH⁻ Kb = 3.7x10⁻⁴
i) x 0 0
e) x - y y y
Now, let's write the expression for the Kb:
Kb = [CH₃NH₃⁺] [OH⁻] / [CH₃NH₂]
We can get the concentrations of the products, because we already know the value of the pH. from there, we calculate the value of pOH and then, the OH⁻:
The pOH:
pOH = 14 - pH
pOH = 14 - 10.68 = 3.32
The [OH⁻]:
[OH⁻] = 10^(-pOH)
[OH⁻] = 10^(-3.32) = 4.79x10⁻⁴ M
With this concentration, we replace it in the expression of Kb, and then, solve for the concentration of methylamine:
3.7*10⁻⁴ = (4.79*10⁻⁴)² / x - 4.79*10⁻⁴
3.7*10⁻⁴(x - 4.79*10⁻⁴) = 2.29*10⁻⁷
3.7*10⁻⁴x - 1.77*10⁻⁷ = 2.29*10⁻⁷
x = 2.29*10⁻⁷ + 1.77*10⁻⁷ / 3.7*10⁻⁴
x = [CH₃NH₂] = 1.097*10⁻³ M
With this concentration, we calculate the moles in 100 mL:
n = 1.097x10⁻³ * 0.100 = 1.097x10⁻⁴ moles
Finally to get the mass, we need to molar mass of methylamine which is 31.05 g/mol so the mass:
m = 1.097x10⁻⁴ * 31.05
<h2>
m = 0.0034 g or 3.4 mg of Methylamine</h2>
