Determine the pH at the equivalence (stoichiometric) point in the titration of 37 mL of 0.1 M

Question

Determine the pH at the equivalence (stoichiometric) point in the titration of 37 mL of 0.1 M (CH₃)₃N(aq) with 0.26 M HCl(aq). The K_b of (CH₃)₃N is 6.5 x 10^-5.

The answer is 5.48

Answer 1

K_b = 6.5 x 10^-5

K_a = (1 x 10^-15)/(6.5 x 10^-5) = 1.54 x 10^-11

pK_a = -log(1.54 x 10^-11) = 10.81

m₁v₁=m₂v₂ (.1M)(37mL) = (.26M)v₂ v₂ = 3.7/.26M = 14.23mL

14.23 is the volume of HCl required to reach the equivalence point.

Total Volume = 37 +14.23 = 51.23mL = .05123L

37/1000 x .1M = 0.0037moles

0.0037moles/.05123L = .0722M

Substitute the known data: (1 x 10^-15)/(6.5 x 10^-5) = x²/.0722 = 1.54 x 10^-11= x²/.0722 = x²=(1.54 x 10^-11) (.0722) = 1.11 x 10^-12, now take the square root of 1.11 x 10^-12 and get x = 1.05 x 10^-6

The pH can be calculated by the following formula:

pH = pKa + log [Base/Acid]

pH = 10.81 +log [Base/Acid]

pH = -log[H⁺]

pH= -log (1.05 x 10^-6) = 5.98