Question

For each strong base solution, determine [H3O+], [OH−], pH, and pOH. in part A,B,and C

A.) 8.78×10−3 M LiOH Express your answer using three significant figures. Enter your answers numerically separated by a comma. [OH−],[H3O+] = M

B.)5.3×10−4 M Ca(OH)2 Express your answer using two significant figures. Enter your answers numerically separated by a comma. [OH−],[H3O+] = M

C.) 5.3×10−4 M Ca(OH)2 Express your answer to two decimal places. Enter your answers numerically separated by a comma. pH,pOH =

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

Answer 1

A) LiOH_(aq) --> Li⁺_(aq)       +       OH^-_(aq)

Initial 8.78×10⁻³-----------------0---------------0----

Equilibrium (0)-----------------( 8.78×10⁻³)-----( 8.78×10⁻³)-

Since it is a strong base it dissociates completely

And concentration of [OH^-] = 8.78×10⁻³M

Now Kw = [H⁺][OH^-] Or Kw = [H₃O⁺][OH^-]

Kw= 1 × 10^-14 = [H₃O⁺] ×8.78×10⁻³M

[H₃O⁺] = 1 × 10^-14/8.78×10⁻³

[H₃O⁺] = 1.1389×10^-12M

pOH = -log [OH^-] = -log 8.78×10⁻³M

pOH =2.0565

B) Ca(OH)_2(aq)               à      Ca²⁺_(aq)       +       2OH^-_(aq)

Initial 5.3×10⁻⁴ M -----------------0---------------------------------0----

Equilibrium (0)----------------------( 5.3×10⁻⁴)-------------------2*( 5.3×10⁻⁴)

Since it is a strong base it dissociates completely

And concentration of [OH^-] = 2 × 5.3×10⁻⁴= 1.06×10⁻³M

Now Kw = [H⁺][OH^-] Or Kw = [H₃O⁺][OH^-]

Kw= 1 × 10^-14 = [H₃O⁺] ×1.06×10⁻³M

[H₃O⁺] = 1 × 10^-14/1.06×10⁻³M

[H₃O⁺] = 9.43×10^-12M

pOH = -log [OH^-] = -log 1.06×10⁻³M

pOH =2.97

C) Na⁺_(aq)+ F^-_(aq)+ H₂O_(l)à NaOH_(aq)+ HF_(aq)----A strong base and a weak acid.

Now we know that Ka × Kb = 1x10^-14

Therefore, Kb = 1x10^-14 / 3.5x 10^-5 = 2.85 x10^-10

  
But, Kb = [Na+][OH-] / [NaF]

Substituting the values,
Kb = x²/ 0.338

2.85 x10^-10= x²/ 0.338

x = 9.19 x10^-6
pOH = -log9.19 x10^-6 = 5.036

  pH = 14-5.036 = 8.96

pH= 8.96