Calculate the pH of the base solution after the chemist has added 5.7 mL of the HNO3 solution to it.

Question

An analytical chemist is titrating 63.6 mL of a 0.1800M solution of diethylamine ((C₂H₅)₂NH) with a 0.6300M solution of HNO₃ The pK_b of diethylamine is 2.89. Calculate the pH of the base solution after the chemist has added 5.7 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.

Answer 1

(C₂H₅)₂NH + HNO₃ ==> (C₂H₅)₂NH₂⁺ + NO₃^-

Initial moles (C₂H₅)₂NH = 63.6 ml x 1 L / 1000 ml x 0.1800 mol / L = 0.01145 mols

Initial moles HNO₃ = 5.7 mls x 1 L / 1000 ml x 0.6300 mol / L = 0.003591 mols

This solution creates a BUFFER consisting of a weak base (C₂H₅)₂NH) and the conjugate acid.

(C₂H₅)₂NH + H⁺ ==> (C₂H₅)₂NH₂⁺

0.01145.....0.003591..................0...........Initial

-0.003591.....-0.003591....+0.003591......Change

0.007859...........0...............0.003591.....Equilibrium

From the Henderson Hasselbalch equation for a basic buffer we have...

pOH = pKb + log [salt] / [base]

pOH = 2.89 + log (0.003591) / (0.007859)

pOH = 2.89 -0.34

pOH = 2.55

pH = 14 - pOH

pH = 11.45