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

Step by step solution

01

Reaction a: Relationship between \(K_a\), \(K_b\), and the equilibrium constant

For reaction a, we are given the equilibrium constant for the amino group, \(K_b\). Since this reaction represents a proton transfer from the carboxylic acid group to the amino group, we can relate these equilibrium constants by: $$ K = \frac{1}{K_b} $$ Thus, we can calculate the equilibrium constant for reaction a as: $$ K = \frac{1}{6.0 \times 10^{-5}} $$

02

Calculate the equilibrium constant for reaction a

Now let's compute the equilibrium constant for reaction a: $$ K = \frac{1}{6.0 \times 10^{-5}} = 1.67 \times 10^{4} $$

03

Reaction b: Relationship between \(K_a\), \(K_b\), and the equilibrium constant

For reaction b, we are given the equilibrium constant for the carboxylic acid group, \(K_a\). Since this reaction represents deprotonation of the carboxylic acid group, we can relate these equilibrium constants by: $$ K = K_a $$ Thus, we can calculate the equilibrium constant for reaction b as: $$ K = 4.3 \times 10^{-3} $$

04

Reaction c: Relationship between \(K_a\), \(K_b\), and the equilibrium constant

For reaction c, we need to relate the given equilibrium constants to the desired equilibrium constant. We can do this by combining reactions a and b: 1. Multiply reaction a by 2: \(2H^{+}NH_3CH_2CO_2^{-} + 2H_2O \rightleftharpoons 2H_2NCH_2CO_2^{-} + 2H_3O^{+}\) 2. Add reaction b: \(H_2NCH_2CO_2^{-} + H_2O \rightleftharpoons H_2NCH_2CO_2H + OH^{-}\) Sum: \(H^{+}H_3NCH_2CO_2H \rightleftharpoons 2H^{+} + H_2NCH_2CO_2^{-}\) The equilibrium constants for these reactions multiply as follows: $$ K = 2 \times (K_a)(K_b) $$

05

Calculate the equilibrium constant for reaction c

Now let's compute the equilibrium constant for reaction c: $$ K = 2 \times (4.3 \times 10^{-3})(6.0 \times 10^{-5}) = 5.16 \times 10^{-7} $$ Putting it all together, the equilibrium constants for reactions a, b, and c are: $$ K_{a} = 1.67 \times 10^{4},\ K_{b} = 4.3 \times 10^{-3},\ K_{c} = 5.16 \times 10^{-7} $$

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