A chemist titrates of a methylamine solution with HBr solution at . Calculate the pH at equivalence. The of methylamine is . Round your answer to decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of solution added.

pH = 5.796

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