Question

What is the pH at the equivalence point for the titration of 25.00 mL 0.100 M CH3CH2CH2CO2- with 0.1848 M HCl? Species (K values) CH3CH2CH2CO2H (Ka = 1.14E-5) CH3CH2CH2CO2- (Kb = 2.63x10-10

Answer 1

The acid-base neutralization reaction is given as

CH₃CH₂CH₂CO₂^- (aq) + HCl (aq) --------> CH₃CH₂CH₂COOH (aq) + Cl^- (aq)

As per the stoichiometric equation,

1 mole CH₃CH₂CH₂CO₂^- = 1 mole HCl = 1 mole CH₃CH₂CH₂COOH

Millimoles of CH₃CH₂CH₂CO₂^- = (25.00 mL)*(0.100 M) = 2.500 mmole = millimoles of HCl required to reach the equivalence point = millimoles of CH₃CH₂CH₂CO₂H formed at the equivalence point.

Milliliters of HCl required to reach the equivalence point = (2.500 mmole)/(0.1848 M) = 13.5281 mL ≈ 13.53 mL.

Total volume of the solution at the equivalence point is (25.00 + 13.53) mL = 38.53 mL and the concentration of CH₃CH₂CH₂COOH is (2.50 mmole)/(38.53 mL) = 0.064884 M ≈ 0.065 M.

At the equivalence point, we have only CH₃CH₂CH₂CO₂H which is a weak acid and undergoes dissociation as

CH₃CH₂CH₂CO₂H (aq) --------> CH₃CH₂CH₂CO₂^- (aq) + H⁺ (aq)

K_a = [CH₃CH₂CH₂CO₂^-][H⁺]/[CH₂CH₂CH₂CO₂H]

=====> 1.14*10^-5 = (x).(x)/(0.065 – x)

Since [CH₃CH₂CH₂CO₂H] is low and K_a is low, we can assume x << 0.065 M and hence,

1.14*10^-5 = x²/(0.065)

====> x² = 7.41*10^-7

====> x = 8.608*10^-4

Therefore, [H⁺] = 8.608*10^-4 M and pH = -log [H⁺] = -log (8.608*10^-4) = 3.065 (ans).