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

Answer:

The answer is "10.2".

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:  

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

$pH= pK_{a_2} + \log \frac{A^{2-}}{HA^{-}}$

      $= 9.60 + \log \frac{0.016}{0.004} \\\\ = 10.2$