In glycine, the carboxylic acid group has \(K_{\mathrm{a}}=4.3 \times 10^{-3}\) and the amino group has \(K_{b}=6.0 \times 10^{-5} .\) Use these equilibrium constant values to calculate the cquilibrium constants for the following. a. \(^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\) b. \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{OH}^{-}\) c. \(\mathbf{H}_{3} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}\)

First, let's define the equilibrium constants we already know: - Ka is the acid dissociation constant for the carboxylic acid group of glycine - Kb is the base dissociation constant for the amino group of glycine For the first reaction, we can treat it as a simple acid dissociation reaction: \(^{+} \mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{3} \mathrm{O}^{+}\) In this reaction, the amino group of glycine is losing a proton. Therefore, Ka for this reaction would be equal to the given Ka value. So, \(K_{a1} = K_{a} = 4.3 \times 10^{-3}\).

For the second reaction, it can be treated as a simple base dissociation reaction: \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{OH}^{-}\) In this reaction, the carboxylic acid group of glycine is gaining a proton. Therefore, Kb for this reaction would be equal to the given Kb value. So, \(K_{b1} = K_{b} = 6.0 \times 10^{-5}\).

The third reaction is as follows: \(\mathrm{H}_{3} \mathrm{NCH}_{2} \mathrm{CO}_{2} \mathrm{H} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CO}_{2}^{-}\) This reaction can be considered as a combination of two reactions: the first given reaction in the reverse direction, and the second given reaction in the forward direction. This means that the equilibrium constant for this reaction can be calculated by multiplying the inverse of the equilibrium constant for the first reaction (Ka) with the equilibrium constant for the second reaction (Kb). Hence, for the third reaction, the equilibrium constant can be calculated as follows: \(K_{c} = \frac{1}{K_{a1}} \times K_{b1} = \frac{1}{4.3 \times 10^{-3}} \times 6.0 \times 10^{-5}\) Calculation: \(K_{c} = 1.3953 \times 10^{-2}\)