Question

1. Using the given information answer the following questions with corresponding answers.

(a)

Serine is a diprotic acid (K_a1 = 6.17 $LaTeX: \times$× 10^-3 and K_a2 = 7.08 $LaTeX: \times$× 10^-10) that can have three different forms in solution: H₂S⁺, HS, and S^- (Note: S is not sulfur in these formulas). What would be the concentration of H₂S⁺ in the solution of 0.105M Na⁺S^-?

Group of answer choices

8.29 $LaTeX: \times$× 10^-8 M

1.62 $LaTeX: \times$× 10^-12 M

9.62 $LaTeX: \times$× 10^-9 M

4.15 $LaTeX: \times$× 10^-11 M

7.08 $LaTeX: \times$× 10^-10 M

(b)

Serine is a diprotic acid (K_a1 = 6.17 $LaTeX: \times$× 10^-3 and K_a2 = 7.08 $LaTeX: \times$× 10^-10) that can have three different forms in solution: H₂S⁺, HS, and S^- (Note: S is not sulfur in these formulas). What would be the isoelectric point of 0.010M serine in solution?

Group of answer choices

6.925

5.680

8.126

7.000

4.152

(c)

Lysine is a triprotic acid (K_a1 = 6.61 $LaTeX: \times$× 10^-3, K_a2 = 1.12 $LaTeX: \times$ × 10^-9, and K_a3 = 2.95 $LaTeX: \times$×10^-11) that can have four different forms in solution: H₃L²⁺, H₂L⁺, HL, and L^-. What would be the pH of the solution containing 0.080M H₃L⁺?

Group of answer choices

1.70

3.08

2.39

2.05

0.73

(d)

Lysine is a triprotic acid (K_a1 = 6.61 $LaTeX: \times$× 10^-3, K_a2 = 1.12 $LaTeX: \times$ × 10^-9, and K_a3 = 2.95 $LaTeX: \times$×10^-11) that can have four different forms in solution: H₃L²⁺, H₂L⁺, HL, and L^-. What would be the concentration of L^- in the solution of 0.080M H₃L⁺?

Group of answer choices

1.1 $LaTeX: \times$× 10^-9 M

2.9 $LaTeX: \times$× 10^-11 M

1.7 $LaTeX: \times$× 10^-18 M

6.7 $LaTeX: \times$× 10^-16 M

4.7 $LaTeX: \times$× 10^-17 M

Answer 1

Q1(a). Correct answer is : 1.62 x 10^-12

Q1(b). Correct answer is : 5.680

Q1(c). Correct answer is : 1.70

Q1(d). Correct answer is : 2.9 x 10^-11 M

Explanation

Q1. The hydrolysis equation for S^- is

S^- (aq) + H₂O (l) $\rightarrow$ HS (aq) + OH^- : K_b1

HS (aq) + H₂O (l) $\rightarrow$ H₂S⁺ (aq) + OH^- : K_b2

K_b2 = (K_w) / (K_a1)

K_b2 = (1.0 x 10^-14) / (6.17 x 10^-3)

K_b2 = 1.62 x 10^-12

For second hydrolysis, concentration of H₂S⁺ is almost equal to K_b2

[H₂S⁺] = K_b2 = 1.62 x 10^-12