A.) In the laboratory, a general chemistry student measured the pH of a 0.497 M

Question

A.) In the laboratory, a general chemistry student measured the pH of a 0.497 M aqueous solution of acetic acid to be2.542.

Use the information she obtained to determine the Ka for this acid.
Ka(experiment) =

B.) In the laboratory, a general chemistry student measured the pH of a 0.497 M aqueous solution of hydrofluoric acidto be 1.740.

Use the information she obtained to determine the Ka for this acid.
Ka(experiment) =

C.) In the laboratory, a general chemistry student measured the pH of a 0.497 M aqueous solution of acetic acid to be2.508.

Use the information she obtained to determine the Ka for this acid.
Ka(experiment) =

D.) The hydroxide ion concentration of an aqueous solution of 0.497 M acetic acidis

[OH-] =  M.

E.) The hydronium ion concentration of an aqueous solution of 0.497 Mhydrofluoric acid (Ka = 7.20×10-4) is

[H3O+] =  M.

F.)The pOH of an aqueous solution of0.497 M acetic acid is

Explanation / Answer

The Ionization of acetic acid is represented as

a) CH3COOH <--------> CH3COO- +H+

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

Given pH= 2.542 , pH= -log[H+] =2.542

[H+] =10(-2.542) =0.002871= [CH3COO-]

At Equilibrium [CH3COOH]=0.497-0.002871 =0.494

Ka= 0.002871*0.002871/ 0.494=1.668*10-5

b

HF<--> H+ F-

pH= 1.740

[H+] =10(-1.740) =0.0182 =[F-]

Ka= [H+] [F-] /[HF] = 0.0182*0.0182/ (0.497-0.0182)=0.000692

C)

The Ionization of acetic acid is represented as

CH3COOH -<-------> CH3COO- +H+

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

Given pH= 2.508 , pH= -log[H+] =2.508

[H+] =10(-2.542) =0.003105= [CH3COO-]

At Equilibrium [CH3COOH]=0.497-0.003105 =0.494

Ka= 0.003105*0.003105/ 0.494=1.951*10-5

d)

CH3COOH <--------> CH3COO- +H+

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

Let x= [H+] =[CH3COO-]

Ka= 1.75*10-5 = x2/(0.497-x)

When solved x= 0.00295, pH=2.53, pOH= 14-pH= 14-2.53=11.47

[OH-] = 10 (-11.47) =3.388*10-12

e)

HF+H2O<---> H3O+ + F-

Ka= [H3O+] [F-] /[HF] = x2/(0.497-x)= 7.2*10-4

Where x= concentration of [H3O+] at equilibrium

When solved for x, the above equation gives [H3O+] =0.0186

f)

CH3COOH -<-------> CH3COO- +H+

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

Let x= [H+] =[CH3COO-]

Ka= 1.75*10-5 = x2/(0.497-x)

When solved x= 0.00295, pH=2.53,

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