Question

Use the given information to answer the following questions with corresponding answers.

(a)

Consider the following dissociation reactions of glycine that can have three different forms in solution: H₂G⁺, HG, or G^-. :

H₂G⁺   $LaTeX: \Longleftrightarrow$ ⟺   HG + H⁺ K_a1 = 4.6 $LaTeX: \times$ × 10^-3

HG   $LaTeX: \Longleftrightarrow$ ⟺   G^- + H⁺   K_a2 = 2.5 $LaTeX: \times$ × 10^-10

Determine the dissociation constant for G^- + H₂O   $LaTeX: \Leftrightarrow$ ⇔   HG + OH^-.

Group of answer choices

4.0 $LaTeX: \times$ ×× 10^-9

4.0 $LaTeX: \times$ ×× 10^-5

2.2 $LaTeX: \times$ ×× 10^-12

3.5 $LaTeX: \times$ ×× 10^-6

2.9 $LaTeX: \times$ ×× 10^-8

^(b)

Consider the following dissociation reactions of phosphate (PO₄^3-):

PO₄^3- + H₂O   $LaTeX: \Leftrightarrow$ ⇔   HPO₄^2- + OH^-   pK_b1 = 1.65

HPO₄^2- + H₂O   $LaTeX: \Leftrightarrow$ ⇔   H₂PO₄^- + OH^-   pK_b2 = 6.98

H₂PO₄^- + H₂O   $LaTeX: \Leftrightarrow$ ⇔   H₃PO₄ + OH^-   pK_b3 = 11.85

Determine the dissociation constant for H₂PO₄^-    $LaTeX: \Leftrightarrow$ ⇔   HPO₄^2- + H⁺.

Group of answer choices

1.4 $LaTeX: \times$ ×× 10^-12

7.1 $LaTeX: \times$ ×× 10^-3

1.0 $LaTeX: \times$ ×× 10^-7

4.5 $LaTeX: \times$ ×× 10^-13

9.5 $LaTeX: \times$ ×× 10^-8

^(c)

Alanine is a diprotic acid (K_a1 = 4.57 $LaTeX: \times$ × 10^-3 and K_a2 = 2.04 $LaTeX: \times$ × 10^-10) that can have three different forms in solution: H₂A⁺, HA, and A^-. What would be the pH of the solution containing 0.085M H₂A⁺Cl^-?

Group of answer choices

3.59

2.11

4.24

2.79

3.05

(d)

Alanine is a diprotic acid (K_a1 = 4.57 $LaTeX: \times$ × 10^-3 and K_a2 = 2.04 $LaTeX: \times$ × 10^-10) that can have three different forms in solution: H₂A⁺, HA, and A^-. What would be the concentration of A^- in the solution of 0.0725M H₂A⁺Cl^-?

Group of answer choices

2.04 $LaTeX: \times$ × 10^-10 M

4.57 $LaTeX: \times$ × 10^-3 M

5.38 $LaTeX: \times$ × 10^-11 M

1.48 $LaTeX: \times$ × 10^-6 M

3.15 $LaTeX: \times$ × 10^-8 M

Answer 1

(a) H₂G⁺ <========> HG + H⁺             K_a1 = [HG][H⁺]/[H₂G⁺] = 4.6*10^-3

HG <=======> G^- + H⁺                         K_a2 = [G^-][H⁺]/[HG] = 2.5*10^-10

Add the auto-ionization reaction of water as below.

H₂O <=======> H⁺ + OH^-                      K_w = [H⁺][OH^-] = 1.0*10^-14

The given reaction is

G^- + H₂O <======> HG + OH^-

The equilibrium constant is written as

K = [HG][OH^-]/[G^-] (the concentration of H₂O is assumed constant)

= ([HG]/[H⁺][G^-])*([H⁺][OH^-])

= K_w/K_a2

= (1.0*10^-14)/(2.5*10^-10)

= 4.0*10^-5 (ans).

(b) The pK_b values are given; determine the K_b values as

pK_b = -log (K_b)

======> K_b = antilog (-pK_b)

pKb	Kb
1.65	2.24*10-2
6.98	1.05*10-7
11.85	7.08*10-11

PO₄^3- + H₂O <======> HPO₄^2- + OH^-                   K_b1 = [HPO₄^2-][OH^-]/[PO₄^3-] = 2.24*10^-2

HPO₄^2- + H₂O <=======> H₂PO₄^- + OH^-              K_b2 = [H₂PO₄^-][OH^-]/[HPO₄^2-] = 1.05*10^-7

H₂PO₄^- + H₂O <=======> H₃PO₄ + OH^-               K_b3 = [H₃PO₄][OH_-]/[H₂PO₄^-] = 7.08*10^-11

Add the auto-ionization reaction of water as below.

H₂O <=======> H⁺ + OH^-                      K_w = [H⁺][OH^-] = 1.0*10^-14

The given reaction is

H₂PO₄^- <=======> HPO₄^2- + H⁺

The equilibrium constant is written as

K = [HPO₄^2-][H⁺]/[H₂PO₄^-]

= ([HPO₄^2-]/[H₂PO₄^-][OH^-])*([H⁺][OH^-])

= K_w/K_b2

= (1.0*10^-14)/(1.05*10^-7)

= 9.52*10^-8

≈ 9.5*10^-8