Question

A.) Find the pH of a 0.338 M NaF solution. (The Ka of hydrofluoric acid, HF, is 3.5×10?4.)

B.) Determine the [OH?] of a 0.30 M solution of NaHCO3. and Determine the pH of this solution.

C.)Find the [OH?] of a 0.46 M  methylamine (CH3NH2) solution. (The value of Kb for methylamine (CH3NH2) is 4.4×10?4.) include units

D.) Find the pH of a 0.46? M  methylamine (CH3NH2) solution.

Answer 1

A)

[NaF] = [F-] = 0.338 M

F- + H2O <==> HF + OH-

let x amount hydrolyzed

Kb = Kw/Ka = [HF][OH-]/[F-]

1 x 10^-14/3.5 x 10^-4 = x^2/0.338

x = [OH-] = 3.11 x 10^-6 M

[H+] = 1 x 10^-14/3.11 x 10^-6 = 3.22 x 10^-9 M

pH = -log[H+] = 8.49

B)

[NaHCO3] = [HCO3-] = 0.3 M

HCO3- + H2O <==> H2CO3 + OH-

let x amount hydrolyzed

Kb = Kw/Ka = [H2CO3][OH-]/[HCO3-]

1 x 10^-14/4.3 x 10^-7 = x^2/0.3

x = [OH-] = 8.35 x 10^-5 M

[H+] = 1 x 10^-14/8.35 x 10^-5 = 1.2 x 10^-10 M

pH = -log[H+] = 9.92

C)

[CH3NH2] = 0.46 M

CH3NH2 + H2O <==> CH3NH3+ + OH-

let x amount hydrolyzed

Kb = [CH3NH3+][OH-]/[CH3NH2]

4.4 x 10^-4 = x^2/0.46

x = [OH-] = 0.014 M

D) For CH3NH2 in C)

[H+] = 1 x 10^-14/0.014 = 7.03 x 10^-13 M

pH = -log[H+] = 12.15