Five Part question (I know its hard but if you can\'t do it, don\'t) Adipic acid

Question

Five Part question (I know its hard but if you can't do it, don't)

Adipic acid, HOOC(CH2)4COOH, is among the top 50 manufactured chemicals in the United States (nearly 1 million metric tons annually). Its chief use is in the manufacture of nylon. It is a diprotic acid having Ka1=3.9×10?5 and Ka2=3.9×10?6.

A saturated solution of adipic acid is about 0.10 M HOOC(CH2)4COOH.

Part A

Calculate the concentration of [H3O+] in this solution.

Express your answer to two significant figures and include the appropriate units.

Part B

Calculate the concentration of [HOOC(CH2)4COO?] in this solution.

Express your answer to two significant figures and include the appropriate units.

Part C

Calculate the concentration of [HOOC(CH2)4COOH] in this solution.

Express your answer to two significant figures and include the appropriate units.

Part D

Calculate the concentration of [?OOC(CH2)4COO?] in this solution.

Express your answer to two significant figures and include the appropriate units.

Part E

Calculate the concentration of [OH?] in this solution.

Express your answer to two significant figures and include the appropriate units.

Part c Calculate the concentration of [HOOC(CH2)4 COOH in this solution. Express your answer to two significant figures and include the appropriate units. the concentration of HOOC(CH).COOHl in this solution. HooC(CH' 0111- Value 1 Units | Value [HOOC(CH2),COOH] = Units Submit My Answers Give Up Part D Calculate the concentration of OOC(CH2)4CO0 in this solution Express your answer to two significant figures and include the appropriate units. r ooCICH),c00 alue Units ValueUnits Units Submit My Answers Give UD

Explanation / Answer

ANSWER:

Lets represent adipic acid as H2A for the sake of simplicity.

H2A <-----------> H+ + HA-   Ka1 = 3.9 X 10-5

Initial 0.1 0 0

Change 0.1-x +x +x

Eq 0.1-x x x

3.9 X 10-5 = x*x/0.1-x   

x2 = 3.9 X 10-5 * 0.1 Neglecting x in dnomenator because x<<0.1

x = x1 = 0.00197M lets designate x as x1

HA- dissociates again as

  HA- <-------------> H+ + A- Ka2 = 3.9 X 10-6

initial 0.00197 0 0

change 0.00197 -x +x +x

Eq   0.0019 -x x x

3.9 = x2/0.00197

x2 = 0.00197 * 3.9 X 10-6    Explaned above

x = x2 = 8.77 X 10-5   lets designate x as x2

A) Total H+ (or H3O+) = x1 + x2 = 0.00197 + 8.77 X 10-5 = 0.00205M

B) concentration of HOOC(CH2)4COO (HA-) = x1 = 0.00197M

C) concentration of HOOC(CH2)4COOH (H2A) = 0.1 - x1 = 0.098M

D)concentration of OOC(CH2)4COO = x2 = 8.77 X 10-5 M

E) [H+][OH-] = 10-14

[OH-] = 10-14 / [H+] = 10-14 / 0.00205 = 4.87X10-12 M

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