A. Calculate the pH of 1.13 MCH3CO2H. B. Calculate the concentration of H3O+ in

Question

A. Calculate the pH of 1.13 MCH3CO2H.

B. Calculate the concentration of H3O+ in 1.13 MCH3CO2H

C. Calculate the concentration of CH3CO2 in 1.13 MCH3CO2H.

D. Calculate the concentration of CH3CO2H in 1.13 MCH3CO2H

E. Calculate the concentration of OH in 1.13 MCH3CO2H.

F. Calculate the pH of 0.0127 MCH3CO2H.

G. Calculate the concentration of H3O+ in 0.0127 MCH3CO2H.

H. Calculate the concentration of CH3CO2 in 0.0127 MCH3CO2H.

I. Calculate the concentration of CH3CO2H in 0.0127 MCH3CO2H.

J. Calculate the concentration of OH in 0.0127 MCH3CO2H.

Explanation / Answer

A. [CH3COOH] = 1.13 M

CH3COOH is a weak acid. So, it will not completely dissociate.

CH3COOH ----> CH3COO- + H+

Ka of CH3COOH = 1.76*10-5

Ka = [CH3COO-][H+]/[CH3COOH]

1.76*10-5 = x.x/(1.13-x)

As dissociation is very less, so 1.13-x = 1.13

1.76*10-5 = x.x/1.13

1.76*10-5 = x2/1.13

x2 = 1.99*10-5

x = 4.46*10-3 M

x = [H+] = 4.46*10-3 M

pH = -log[H+] = -log(4.46*10-3) = 2.35

pH = 2.35

B) [H3O+] = [H+] = 4.46*10-3 M

C) [CH3COO-] = x = 4.46*10-3 M

D) [CH3COOH] = 1.13-x = 1.13 - (4.46*10-3 M) = 1.13 M (approx.)

E) pH = 2.35

pOH = 14-pH = 14-2.35 = 11.65

pOH = -log[OH-]

11.65 = -log[OH-]

[OH-] = 2.24*10-12 M

F) [CH3COOH] = 0.0127 M

CH3COOH is a weak acid. So, it will not completely dissociate.

CH3COOH ----> CH3COO- + H+

Ka of CH3COOH = 1.76*10-5

Ka = [CH3COO-][H+]/[CH3COOH]

1.76*10-5 = x.x/(0.0127-x)

As dissociation is very less, so 0.0127-x = 0.0127

1.76*10-5 = x.x/0.0127

1.76*10-5 = x2/0.0127

x2 = 2.24*10-7

x = 4.73*10-4 M

x = [H+] = 4.73*10-4 M

pH = -log[H+] = -log(4.73*10-4) = 3.33

pH = 3.33

G) [H3O+] = [H+] = 4.73*10-4 M

H) [CH3COO-] = x = 4.73*10-4 M

I) [CH3COOH] = 0.027-x = 0.027 - (4.73*10-4 M) = 0.027 M (approx.)

J) pH = 3.33

pOH = 14-pH = 14-3.33 = 10.67

pOH = -log[OH-]

10.67 = -log[OH-]

[OH-] = 2.14*10-11 M

[CH3COOH] [CH3COO-] [H+] Initial 1.13 M 0 0 Change -x +x +x Equilibrium 1.13-x x x

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