Question

7. Calculate the pH of the following salt solutions

a) 0.388 M LiC₂H₃O₂

b) 0.369 M CH₃NH₃NO₃

c) 0.194 M LiClO₄

Answer 1

Q.7

a)

            LiC2H3O2 --- > C2H3O2^- + Li+                 

C2H3O2^- is conjugate base of acetic acid.

Reaction of it with water and ICE

            C2H3O2^- + H2O --- > C2H3O2H + OH^-

I           0.388                           0                      0

C         -x                                 +x                    +x

E          (0.388-x)                     x                      x

Kb expression

Kb = [C2H3O2H][OH-]/[ C2H3O2^-]

We can get kb value by using ka of acetic acid

Ka = 1.8 E-5

Kb = 1.0 E-14 / ka = 1.0 E-14 / 1.8 E-5

Plug kb and equilibrium concentration in kb expression .

5.55556E-10 =x²/ (0.388- x)

Since the value of kb is very small we can neglect x in the denominator.

5.55556E-10 =x²/ 0.388

x =[ OH-]= 1.46818E-05        M

Calculate pOH

pOH = - log [OH-]

            = - log (5.55556E-10)

pOH=4.83322039      

pH = 14 - pOH

pH=14-4.833

pH=9.2                                   

            b)                    

            [CH3NH3NO3]= 0.369 M

            [CH3NH3+] = 0.369   M

CH3NH3⁺ is conjugate acid of methyl amine

            kb methyl amine = 4.38E-04

            ka =1.0 E-14 / 4.38 E-4

            ka=2.28E-11  

Reaction of conjugate acid with water and ICE

            CH3NH3⁺ + H2O -- > CH3NH2 + H3O⁺

I           0.369                           0                      0

C         -x                                 +x                    +x

E          (0.369-x)                     x                      x

Ka expression

Ka = 2.28 E-11 = x²/ 0.369-x

Since the value of ka is very small we can neglect x in the denominator.

            2.28 E-11 = x²/ 0.369-x

Lets solve for x

X=[H3O⁺] = 2.90253E-06 M

pH = -log ( 2.90253E-06)

pH= 5.54

                                   

            c)                    

            LiClO4] =0.194          M

            [ClO4^-]= 0.194            M

            pH =7

ClO4^- is the conjugate base of strong acid so it will not make any change in pH.

The pH of the solution is 7 since strong acid strong base reaction has neutral pH.