The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×105. Part A Ca

Question

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×105.

Part A

Calculate the equilibrium concentration of H3O+ in the solution if the initial concentration of C6H5COOH is 6.6×102 M .

Express your answer using two significant figures.

SubmitRequest Answer

Part B

Calculate the equilibrium concentration of C6H5COO in the solution if the initial concentration of C6H5COOH is 6.6×102 M .

Express your answer using two significant figures.

SubmitRequest Answer

Part C

Calculate the equilibrium concentration of C6H5COOH in the solution if the initial concentration of C6H5COOH is 6.6×102 M .

Express your answer using two significant figures.

Please help!!!! I can't figure these out at all :(

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×105.

Part A

Calculate the equilibrium concentration of H3O+ in the solution if the initial concentration of C6H5COOH is 6.6×102 M .

Express your answer using two significant figures.

[H3O+] =   M  

SubmitRequest Answer

Part B

Calculate the equilibrium concentration of C6H5COO in the solution if the initial concentration of C6H5COOH is 6.6×102 M .

Express your answer using two significant figures.

[C6H5COO] =   M  

SubmitRequest Answer

Part C

Calculate the equilibrium concentration of C6H5COOH in the solution if the initial concentration of C6H5COOH is 6.6×102 M .

Express your answer using two significant figures.

Please help!!!! I can't figure these out at all :(

Explanation / Answer

Given,

Benzoic acid Ka = 6.3 x 10^-5

Part A)

[C6H5COOH] initial concentration = 6.6 x 10^-2 M

let x amount of acid dissociated by equation,

C6H5COOH + H2O <==> C6H5COO- + H3O+

So,

Ka = [C6H5COO-][H3O+]/[C6H5COOH]

6.3 x 10^-5 = x^2/(6.6 x 10^-2 - x)

x^2 + 6.3 x 10^-5x - 4.16 x 10^-6 = 0

solving quadratic equation,

x = [H3O+] = 2.01 x 10^-3 M

equilibrium concentration of [H3O+] = 2.01 x 10^-3 M

Part B)

equilibrium concentration of [C6H5COO-] = 2.01 x 10^-3 M

Part C)

equilibrium concentration of [C6H5COOH] = 6.6 x 10^-2 - 2.01 x 10^-3 = 6.4 x 10^-2 M

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.