Question
The amino acid glycine (NH2CH2COOH) is basic because of its –NH2 group and acidic because of its –COOH group. By a process equivalent to base dissociation, glycine can acquire an additional proton to form$\overset{\oplus}{N}H_{3}CH_{2}COOH$. The resulting cation may be considered to be a diprotic acid, since one proton from the
–COOH group and one from the $- \overset{\oplus}{N}H_{3}$ group may be lost. The pKa values for these processes are 2.22 and 9.78, respectively. For a 0.01 M solution of neutral glycine (log 1.7 = 0.22, log 6 = 0.78),
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Solution
$\overset{+}{N}H_{3}CH_{2}COOH\overset{\quad P^{K_{a1}} = 2.22\quad}{\rightarrow}\overset{+}{N}H_{3}CH_{2}COO^{-}\overset{\quad P^{K_{a2}} = 9.78\quad}{\rightarrow}NH_{3}CH_{2}COO^{-}$
$P^{H} = \frac{1}{2}\left( P^{K_{a1}} + P^{K_{a2}} \right) = \frac{1}{2}(2.22 + 9.78) = 6.0$
Now, $K_{a_{1}} = \frac{\left\lbrack \overset{\oplus}{N}H_{3}CH_{2}COO^{-} \right\rbrack\left\lbrack H^{+} \right\rbrack}{\left\lbrack \overset{\oplus}{N}H_{3}CH_{2}COOH \right\rbrack} \Rightarrow 10^{- 2.22} = \frac{0.01 \times 10^{- 6}}{\left\lbrack \overset{\oplus}{N}H_{3}CH_{2}COOH \right\rbrack}$
$\therefore\left\lbrack \overset{\oplus}{N}H_{3}CH_{2}COOH \right\rbrack = 10^{- 5.78} = 1.7 \times 10^{- 6}\text{ M}$
% of glycine in cationic form $= \frac{1.7 \times 10^{- 7}}{0.01} \times 100 = 0.017\%$
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