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 [tex]NaOH \ to\ H_2A[/tex] acid:
Get balance moles as follows:
[tex]\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\\\\[/tex]
In the second equilibrium:
[tex]\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\\\\[/tex]
[tex]pH= pK_{a_2} + \log \frac{A^{2-}}{HA^{-}} \\\\[/tex]
[tex]= 9.60 + \log \frac{0.016}{0.004} \\\\ = 10.2[/tex]
