Question

A 65.0 mL solution of 0.113 M potassium alaninate (H2NC2H5CO2K) is titrated with 0.113 M HCl. The pKa values for the amino acid alanine are 2.344 (pKa1) and 9.868 (pKa2), which correspond to the carboxylic acid and amino groups, respectively. a) Calculate the pH at the first equivalence point b) Calculate the pH at the second equivalence point

Answer 1

Given pKa1 = 2.344

=> Ka2 = 10^-2.344 = 4.53 x 10^-3

Given pKa2 = 9.868

=> Ka2 = 10^-9.868 = 1.355 x 10^-10

(a): For first equivalene point: since the -COOH group has higher pKa value it will be first neutralized to achieve the first equivalence point. Let's represent the diprotic alanite as H2A. The neutralization reaction is

H2A + HCl ---- > HA- + H2O

1 mol, 1 mol ----- 1 mol

Moles of potassium alaniate (H2NC2H5CO2K) taken = MxV = 0.113M x 0.0650 L = 0.007345 mol

Concentration of HCl taken = 0.113 M HCl

Since both HCl and potassium alaniate (H2NC2H5CO2K) have same concentration, the first equivalence point will be achieved after the addition of 65.0 mL ( 0.0650L) of HCl.

Hence total moles of HA- salt formed = 0.007345 mol

Total volume, Vt = 0.065 L + 0.065 L = 0.130 L

Hence [HA-] =  0.007345 mol / 0.130L = 0.0565 M

Hence at equivalence point it will act as a HA- salt and the pH can be calculated from salt hydrolysis. Also weak acid HA- will undergo dissociation. Hence pH will be calculated from the following formulae

(b): Now we again need to add 0.0650 L of HCl to achieve the 2nd equivalent point. and the reaction is

HA- + HCl ----> A2- + H2O

Moles of A2- formed =  0.113M x 0.0650 L = 0.007345 mol

Total volume = 0.0650 L + 0.0650 L + 0.0650 L = 0.195 L

Hence [A2-] =  0.007345 mol / 0.195 L = 0.0377 M

When the second equivalence point is achieved there will be only A^2-. Hence the pH will be due to hydrolysis of A^2- . Hence

Hence pH after second equivalence point will be 11.21