Would the ph be 10.89 at equivalence or after equivalence?

Question

An analytical chemist is titrating 241.3mL of a 0.07900M solution of ethylamine C2H5NH2 with a 0.8500M solution of HIO3. The pKb of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 10.2mL of the HIO3 solution to it.

Answer 1

C₂H₅NH₂ + H⁺ ==> C₂H₅NH₃⁺ ... reaction taking place

initial moles C₂H₅NH₂ = 241.3 ml x 1 L / 1000 ml x 0.07900 mol / L = 0.1906 mols

initial moles H⁺ = 10.2 ml x 1 L / 1000 ml x 0.8500 mol / L = 0.008670 mols

Setting up an ICE tabe:

C₂H₅NH₂ + H⁺ ==> C₂H₅NH₃⁺

0.1906.........0.00867..........0.............Initial

0.00867.......-0.00867.......+0.00867...Change

0.1819...........0.................0.008670....Equilibrium

Using the Henderson Hasselbalch equation to solve for pOH and pH:

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

pOH = 3.19 + log (0.008670 / 0.1819)

pOH = 3.19 + log 0.04766 = 3.19 + (-1.322)

pOH = 1.8682

pH = 12.132

At equivalence you have the mols C₂H₅NH₂ = mols H⁺, meaning all of the C₂H₅NH₂ is converted to C₂H₅NH₃⁺. Since this is the salt of a weak base (C₂H₅NH₂) and a strong acid (HIO₃), the pH of the salt will be acidic (< 7), so a pH of 10.89 would have to be AFTER equivalence.