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Stolb23 [73]
3 years ago
6

A 100.00-mL sample of 0.2000M glycine, A+ form (see structure below), was titrated with 0.2000M of NaOH. Ka1 of glycine = 3.16x1

0-3 ; Ka2 of glycine = 2.51x10-10
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
6 0

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

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