Answer:
pH = 5.796
Step-by-step explanation:
The chemist titrates 110.0mL of a 0.1635M solution of methylamine with HBr 0.3545M. And pKb of methylamine is 3.36
Moles of 110.0mL of the solution of methylamine are:
0.1100L × (0.1635mol / L) = 0.01799 moles of methylamine.
The reaction with HBr is:
CH₃NH₂ + HBr → CH₃NH₃⁺ + Br⁻
At equivalence point, moles of HBr = moles of CH₃NH₂. That means you will have just CH₃NH₃⁺ in solution.
Volume of 0.3545M HBr to reach equivalence point are:
0.01799 moles ₓ (1L / 0.3545mol) = 0.0507L of HBr.
Thus, total volume of solution at equivalence point are:
0.1100L + 0.0507L = 0.1607L
As moles of CH₃NH₃⁺ are equal to moles of methylamine, 0.01799 moles, molarity of the solution is:
0.01799 mol / 0.1607L = 0.1119M
The equilibrium of CH₃NH₃⁺ in water is:
CH₃NH₃⁺ ⇄ CH₃NH₂ + H⁺
Where Ka is defined as:
Ka = [CH₃NH₂] [H⁺] / [CH₃NH₃⁺]
Knowing:
pKa = 14 - pKb,
pKa of methylamine is:
pKa = 14 - 3.36
pKa = 10.64
pKa = -log Ka
Ka = 2.29x10⁻¹¹
Molarity in equilibrium of each species knowing concentration of CH₃NH₃⁺ is 0.1119M are:
[CH₃NH₂] = X
[H⁺] = X
[CH₃NH₃⁺] = 0.1119M - X
Where X is reaction coordinate
Replacing in Ka equation:
2.29x10⁻¹¹ = [X] [X] / [0.1119 - X]
2.29x10⁻¹¹ = [X] [X] / [0.1119 - X]
2.56x10⁻¹² - 2.29x10⁻¹¹X = X²
0 = X² - 2.56x10⁻¹² + 2.29x10⁻¹¹X
Solving for X:
X = -1.6x10⁻⁶ → False solution. There is no negative concentrations.
X = 1.6x10⁻⁶ → Right answer
That means [H⁺] = 1.6x10⁻⁶
as pH = -log [H⁺].
pH = 5.796