1a) The hydronium ion concentration of an aqueous solution of 0.498 M nitrous ac

ID: 524178 • Letter: 1

Question

1a) The hydronium ion concentration of an aqueous solution of 0.498 M nitrous acid (Ka = 4.50×10-4) is [H3O+] =

1b) The pH of an aqueous solution of 0.498 M benzoic acid , C6H5COOH is

1c) The pOH of an aqueous solution of 0.424 M acetylsalicylic acid (aspirin), HC9H7O4, is

1d) The value of Ka for acetic acid is 1.80×10-5. What is the value of Kb, for its conjugate base, CH3COO-?

1e) The value of Ka for acetylsalicylic acid (aspirin) is 3.00×10-4. What is the value of Kb, for its conjugate base, C9H7O4-?

1f) The value of Ka for phenol (a weak acid) is 1.00×10-10. What is the value of Kb, for its conjugate base, C6H5O-?

1a)

nitrous acid is weak so

HNO2 <-> H+ NO2-

Ka = [H+][NO2-]/[HNO2]

(4.5*10^-4) = x*x/(0.498-x)

x = H+ = 0.01475

pH = -log(0.01475) = 1.8312

1b)

pH for C6H5COOH

pKa = 4.20

Ka = 10^-4.20

similarly:

Ka = [H+][Benzoate-]/[HBenzoic]

10^-4.20 = x*x/(0.498-x)

x = H =0.005574

pH = -log(0.005574) = 2.25

1c)

pOH for HC9H7O4

Ka = 3*10^-4

Ka = [H+][Asp-]/[HAsp]

3*10^-4 = x*x/(0.424-x)

x = 0.0111

[H+]= 0.0111

pH = -log(0.0111) =1.9546

pOH = 14-1.9546 = !2.05

d)}

find Kb for acetic acid given

Kw = Ka*Kb

Kb = Kw/Ka = (10^-14)/1.8*10^-5 = 5.55*10^-10

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