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1. Using the given information answer the following questions with corresponding answers. (a) Serine is a diprotic acid...

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

(a)

Serine is a diprotic acid (Ka1 = 6.17 LaTeX: \times× 10-3 and Ka2 = 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 (Ka1 = 6.17 LaTeX: \times× 10-3 and Ka2 = 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 (Ka1 = 6.61 LaTeX: \times× 10-3, Ka2 = 1.12 LaTeX: \times × 10-9, and Ka3 = 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 (Ka1 = 6.61 LaTeX: \times× 10-3, Ka2 = 1.12 LaTeX: \times × 10-9, and Ka3 = 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

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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- : Kb1

HS (aq) + H2O (l) \rightarrow H2S+ (aq) + OH- : Kb2

Kb2 = (Kw) / (Ka1)

Kb2 = (1.0 x 10-14) / (6.17 x 10-3)

Kb2 = 1.62 x 10-12

For second hydrolysis, concentration of H2S+ is almost equal to Kb2

[H2S+] = Kb2 = 1.62 x 10-12

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