Question

A 100.00-mL sample of 0.2000M glycine, A+ form (see structure below), was titrated with 0.2000M of NaOH. Ka1 of glycine = 3.16x10-3 ; Ka2 of glycine = 2.51x10-10

Answer 1

The answer is "10.2".

Step-by-step explanation:

Please find the complete question in the attachment file.

Calculating the pH after adding 180.0 mL of
`NaOH \ to\ H_2A` acid:

Get balance moles as follows:

`\to H_2A + OH^(-) \rightleftharpoons HA^(-) +H_2O \\\\I(mol) \ \ \ \ \ 0.02 \ \ \ \ \ 0.036 \ \ \ \ \ 0 \\\\C(mol) \ \ \ \ \ -0.02 \ \ \ \ \ -0.02 \ \ \ \ \ + 0.02 \\\\E (mol) \ \ \ \ \ \approx 0 \ \ \ \ \ \approx 0.016 \ \ \ \ \ 0.02\\\\`

In the second equilibrium:

`\to HA^(-) + OH^(-) \rightleftharpoons A^(2-) + H_2 O \\\\I(mol) \ \ \ \ \ 0.02 \ \ \ \ \ 0.016 \ \ \ \ \ 0 \\\\C(mol) \ \ \ \ \ -0.016 \ \ \ \ \ -0.016 \ \ \ \ \ + 0.016 \\\\E (mol) \ \ \ \ \ 0.004 \ \ \ \ \ \approx 0 \ \ \ \ \ 0.016\\\\`

`pH= pK_(a_2) + \log (A^(2-))/(HA^(-)) \\\\`

`= 9.60 + \log (0.016)/(0.004) \\\\ = 10.2`