Question

A 65.0 mL solution of 0.118 M potassium alaninate (H2NC2H5CO2K) is titrated with 0.118 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

a) at first equivalence point pH = ( 1/2) ( pka1 + pka2)

hence pH = ( 1/2) ( 2.344+9.868) = 6.1

b) at 2nd equivalence point we have

H2NC2H5CO2K (aq) + 2H+ (aq) ----> +3HNC2H5CO2H (aq)

H2NC2H5CO2K moles = M x V ( in L) = 0.118 x ( 65/1000) = 0.00767

HCl moles = 2 x 2HC2H5CO2K moles = 2 x 0.00767 = 0.01534

to get second equivalence point HCl volume needed = V

Molarity = mmooles/volume

0.118 = 0.01534 /V

V = 0.13 L .

now total solution volume = 0.065+0.13 = 0.195 L

at 2nd equivalence all 2HNC2H5CO2K converted to +3HNC2H5CO2H , now we have reverse equilibrium

+3HNC2H5CO2H (aq) <---> +3HNC2H5COO- (aq) + H+ (aq) , Ka1 = 10^ -pka1 = 10^ -2.344 = 0.004523

at equilibrium we have +3HNC2H5CO2H moles = 0.00767-X , [ +3NC2H5CO2H] = ( 0.00767-X) / 0.195

+3HNC2H5CO2- moles = X = H+ moles

[+3HNC2H5CO2-] =[H+] = X /0.195

Ka = [+3HNC2H5CO2-][H+] /[+3HNC2H5CO2H]

0.00452 = ( X/0.195) ( X/0.195) / ( 0.00767-X) /0.195

0.0008814 = X^2 / ( 0.00767-X)

X^2 + 0.0008814X - 6.76 x 10^ -6 = 0

X = 0.002196 = H+ moles

[H+] = 0.002196/0.195 = 0.01126

pH = -log [H+] = -log ( 0.01126) = 1.95