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

Question

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

A.) 8.78×103 M LiOH Express your answer using three significant figures. Enter your answers numerically separated by a comma. [OH],[H3O+] = M

B.)5.3×104 M Ca(OH)2 Express your answer using two significant figures. Enter your answers numerically separated by a comma. [OH],[H3O+] = M

C.) 5.3×104 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 3.5×104.)

Explanation / Answer

Answer:

A) LiOH(aq) --> Li+(aq)       +       OH-(aq)

Initial 8.78×103-----------------0---------------0----

Equilibrium (0)-----------------( 8.78×103)-----( 8.78×103)-

Since it is a strong base it dissociates completely

And concentration of [OH-] = 8.78×103M

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

Kw= 1 × 10-14 = [H3O+] ×8.78×103M

[H3O+] = 1 × 10-14/8.78×103

[H3O+] = 1.1389×10-12M

pOH = -log [OH-] = -log 8.78×103M

pOH =2.0565

B) Ca(OH)2(aq)               à      Ca2+(aq)       +       2OH-(aq)

Initial 5.3×104 M -----------------0---------------------------------0----

Equilibrium (0)----------------------( 5.3×104)-------------------2*( 5.3×104)

Since it is a strong base it dissociates completely

And concentration of [OH-] = 2 × 5.3×104= 1.06×103M

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

Kw= 1 × 10-14 = [H3O+] ×1.06×103M

[H3O+] = 1 × 10-14/1.06×103M

[H3O+] = 9.43×10-12M

pOH = -log [OH-] = -log 1.06×103M

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

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