Calculate the pH during the titration of 20.00 mL of 0.1000 M ethylamine

Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M ethylamine, C₂H₅NH₂(aq), with 0.2000 M HNO₃(aq) after 3.5 mL of the acid have been added. K_b of ethylamine = 6.5 x 10^-4.

The answer is 11.08

Answer 1

C₂H₅NH₂ + H⁺ ===> C₂H₅NH₃⁺

weak base.................conjugate acid (salt)

This makes a buffer since you have a weak base and the salt of that base.

Using Henderson Hasselbalch equation we can find the pH. For a basic buffer this is...

pOH = pKb + log [salt]/[base]. So, now we must find the [salt] and [base] after addition of HNO3

moles H⁺ added = 3.5 ml x 1 L/1000 ml x 0.2000 mol/L = 0.0007 moles H⁺

Initial moles base = 20.00 ml x 1 L/1000 ml x 0.1000 mol/L = 0.002 moles base

Final moles base = 0.002 - 0.0007 = 0.0013 moles base

Final moles salt (conjugate acid) = 0.0007 moles salt

Final volume = 20 ml + 3.5 ml = 23.5 ml = 0.0235 L

Final [salt] = 0.0007 moles/0.0235 L = 0.0298 M salt

Final [base] = 0.0013 moles/0.0235 L = 0.0553 M base

Substituting into the Henderson Hasselbalch equation and using pKb = -log Kb = -log 6.5x10-4 = 3.19

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

pOH = 3.19 + log (0.0298/0.0553)

pOH = 3.19 + (-0.27)

pOH = 2.02

pH = 14 - pOH

pH = 11.08

You could have done all of the above using an ICE table as follows:

C₂H₅NH₂ + H⁺ ===> C₂H₅NH₃⁺

.002 mol.....0.0007 mol.......0...........Initial

-0.0007......-0.0007.........+0.0007....Change

0.0013mol.....0.................0.0007 mol ...Equilibrium

Then convert moles to molar by dividing by the final volume of 0.0235 L