Question

In glycine, the carboxylic acid group has${\mathbf{K}}_{\mathbf{a}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$and the amino group haslocalid="1664226318911" ${\mathbf{K}}_{\mathbf{b}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$. Use these equilibrium constant values to calculate the equilibrium constants for the following.

a.${}^{\mathbf{\text{+}}}{\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{NCH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{2}}^{\mathbf{-}}{\mathbf{\text{+H}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}\mathbf{\rightleftharpoons }{\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{NCH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}^{\mathbf{\text{-}}}\mathbf{\text{+}}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}$

b.${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{NCH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}^{\mathbf{\text{-}}}{\mathbf{\text{+H}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}\mathbf{\rightleftharpoons }{\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{NCH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}{\mathbf{\text{H + OH}}}^{\mathbf{\text{-}}}$

c.localid="1664232339085" ${}^{\mathbf{\text{+}}}{\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{NCH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}\mathbf{\text{H}}\mathbf{\rightleftharpoons }{\mathbf{\text{2H}}}^{\mathbf{\text{+}}}{\mathbf{\text{+H}}}_{\mathbf{\text{2}}}{\mathbf{\text{NCH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}^{\mathbf{\text{-}}}$

Short answer

The equilibrium constants are:

(a)$K_{eq}=1.7\times {10}^{-10}$

(b) $K_b=2.3\times {10}^{-12}$

(c)$K_{eq}=.3\times {10}^{-13}$

Step-by-step solution

To calculate equilibrium constant for part (a)

First, to be able to solve for the equilibrium constant expression of the following balanced equation,

${}^{\text{+}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_2^{-}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}^{\text{-}}\text{+}{H}_3{O}^{+}\theta$

Since the equation is acidic, we must make use of the given $K_b=6.0\times {10}^{-5}$for the amino group or the conjugate base in the equation.

Then, we write the equilibrium constant expression. In this case, the conjugate base is related to the ${\text{K}}_{\text{b}}{\text{,K}}_{\text{a}}$ or acidity of NH₃. Hence we use the following formula, and solve:

$K_{eq}=\frac{K_w}{K_b}=\frac{1.0\times {10}^{-14}}{6.0\times {10}^{-5}}=1.7\times {10}^{-10}$.

Hence_{$K_{eq}=1.7\times {10}^{-10}$.}

To calculate equilibrium constant for part (b)

First, to be able to solve for the equilibrium constant expression of the following balanced equation,

${\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}{\text{H + OH}}^{\text{-}}$

Since the equation is basic, we must make use of the givenfor the carboxylic acid group or the conjugate acid in the equation.

Then, we write the equilibrium constant expression. In this case, the conjugate acid is related to the $K_a$, thus we must find the $K_b$or basicity of. Hence, we use the following formula, and solve:

$K_b=\frac{K_w}{K_a}=\frac{1.0\times {10}^{-14}}{4.3\times {10}^{-3}}=2.3\times {10}^{-12}$

Hence$K_b$ =$2.3\times {10}^{-12}$.

To calculate equilibrium constant for part (c)

First, to be able to solve for the equilibrium constant expression of the following balanced equation,

${}^{\text{+}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}\text{H}\rightleftharpoons {\text{2H}}^{\text{+}}{\text{+H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}^{\text{-}}$

To solve for the equilibrium constant expression, we must make use of the given$K_a=4.3\times {10}^{-3}$ for the carboxylic group present in the equation and the previous$K_{eq}=1.7\times {10}^{-10}$ for the NH₃ group present in the equation.

Then, we write the equilibrium constant expression. Hence, we use the following formula, and solve:

$K_{eq}=K_a\left(-C{O}_H\right)K_a\left(-N{H}_3^{+}\right)=\left(4.3\times {10}^{-3}\right)\left(1.7\times {10}^{-7}\right)=7.3\times {10}^{-13}$

Thus, the or the equilibrium constant expression for the balanced equation is $7.3\times {10}^{-13}$.

Hence $K_{eq}=7.3\times {10}^{-13}$.