Calculate the pH at the halfway point and at the equivalence point for each of the following:

Question

Calculate the pH at the halfway point and at the equivalence point for each of the following:

b) 100 mL of 0.10 C₂H₅NH₂ (K_b - 5.6 x 10^-4) titrated by 0.20 M HNO₃

I am lost with this process so please explain as simply as possible.

The answers are:

b) 10.74 , 5.96

Please list the steps so I can follow along. Thank you!

Answer 1

C₂H₅NH₂ + H⁺ ==> C₂H₅NH₃⁺

At the halfway point, [C₂H₅NH₂] = [C₂H₅NH₃⁺] and using the Henderson Hasselbalch equation for a base...

pOH = pKb + log [salt]/[base] and pKb = -log Kb = -log 5.6x10-4 = 3.25

pOH = 3.25 + log 1 = 3.25

pH = 14 - 3.25 = 10.75

At equivalence, all the C₂H₅NH₂ is converted to C₂H₅NH₃⁺ so we must consider the hydrolysis of C₂H₅NH₃⁺

moles C₂H₅NH₃⁺ = 100 ml x 1 L/1000 ml x 0.1 mol/L = 0.01 mol

Final volume = 100 ml + 50 ml HNO3 = 150 ml = 0.150 L

Final [C₂H₅NH₃⁺] = 0.01 mol/0.15 L = 0.067 M

C₂H₅NH₃⁺ + H₂O ==> H⁺ + C₂H₅NH₂

Ka = 1x10^-14/Kb = 1x10^-14/5.6x10^-4 = 1.8x10^-9

1.8x10^-11 = (x)(x)/0.067 - x and if we assume x is small we can ignore it in the denominator

x² = 1.2x10^-12

x = 1.09x10-6 = [H+]

pH = -log [H+]

pH = 5.96