Chemistry HW help

Question

A student titrated 30.00 mL of a weak base solution, 0.150 M CH3NH2 (Kb = 4.40 × 10^-4), with 0.100 M HCI solution. Calculate the pH at the equivalence point.

Answer 1

CH3NH2 + HCl ==> CH3NH3Cl

moles CH3NH2 = 30.00 ml x 1 L / 1000 ml x 0.150 mol / L = 0.00450 mols

At equivalence point mols HCl = mols CH3NH2 = 0.00450 mols HCl

The volume of HCl required = 0.00450 mols HCl x 1 L / 0.100 mol = 0.045 L = 45 mls

TOTAL volume at equivalence point = 30.00 ml + 45.00 ml = 75.00 mls = 0.075 L

Also, at equivalence point all CH3NH2 is converted to 0.00450 mols CH3NH3Cl.

To find pH, we need to look at the hydrolysis of 0.00450 mols CH3NH3Cl in 0.0.075 L solution:

[CH3NH3Cl] = 0.00450 mol / 0.075 L = 0.0600 M

CH3NH3⁺ + H2O ==> CH3NH2 + H3O⁺ (CH3NH3⁺ is acting as an acid)

Ka = [CH3NH2][H3O⁺] / [CH3NH3⁺]

Ka = 1x10^-14 / 4.40x10^-4 = 2.27x10^-11

2.27x10^-11 = (x)(x) / 0.0600 - x (assume x is small relative to 0.0600 and ignore it)

x² = 1.36x10^-12

x = 1.17x19^-6 M = [H3O⁺] (note: this is small relative to 0.06 so above assumption was valid)

pH = -log [H3O⁺] = -log 1.17x10^-6

pH = 5.93

(be sure to check the math)