Calculate the pH of the base solution

Question

An analytical chemist is titrating 149.5 mL of a 0.3500 M solution of methylamine (CH₃NH₂) with a 0.6100M solution of HNO₃. The pK_b of methylamine is 3.36. Calculate the pH of the base solution after the chemist has added 102.7 mL of the HNO₃ solution to it.

you may assume the final volume equals the initial volume of the solution plus the volume of HNO, solution added.

Round your answer to 2 decimal places.

pH =

Answer 1

CH₃NH₂ + HNO₃ ==> CH₃NH₃⁺ + NO₃^-

Initial mols CH₃NH₂ = 149.5 ml x 1 L / 1000 ml x 0.3500 mol / L = 0.05233 mols

Initial mols HNO₃ = 102.7 ml x 1 L / 1000 ml x 0.6100 mol / L = 0.06265 mols

NOTE: because there are more mols of HNO₃ than of CH₃NH₂ , HNO₃ will be in excess and the solution

will be acidic

CH₃NH₂ + HNO₃ ==> CH₃NH₃⁺ + NO₃^-

0.05233......0.06265............0..............0..............Initial

-0.05233...-0.05233.......+0.05233...+0.05233....Change

0............0.01032...........0.05233....0.05233....Equilibrium

Final volume = 149.5 ml + 102.7 ml = 252.2 mls = 0.2522 L

Final [HNO₃] = 0.01032 mol / 0.2522 L = 0.04092 M

Final [CH₃NH₃⁺] = 0.05233 mol / 0.2522 L = 0.2075

The final pH will be predominantly dictated by the [HNO3] left over as CH₃NH₃⁺ is such a weak acid as to not contribute significantly to the pH.

pH = -log [HNO₃]

pH = -log 0.04092

pH = 1.39