Question

1.

a)The hydronium ion concentration of an aqueous solution of 0.56 M phenol (a weak acid), C₆H₅OH, is

[H₃O⁺] = ? M

b)The hydronium ion concentration of an aqueous solution of 0.558 M pyridine (a weak base with the formula C₅H₅N) is

[H₃O⁺] = ? M.

2.

a)The pOH of an aqueous solution of 0.445 M acetylsalicylic acid (aspirin), HC₉H₇O₄, is  .?

b)The pH of an aqueous solution of 0.517 M aniline (a weak base with the formula C₆H₅NH₂) is ?.

P.S. Ka values are from the common table

Answer 1

1)

a)

Concentration of aqueous solution of Phenol = C6H5OH = 0.56M

   Ka of Phenol = 1.6 x10^-10

Ka = 1.6 x10^-10

                          C6H5OH + H2O ------------------- C6H5O-    + H3O+

Initial                      0.56                                           0              0

change                     -x                                            +x               +x

equilibrium            0.56-x                                          +x                +x

Ka = [C6H5O-] [ H3O+] / [C6H5OH]

1.6 x10^-10 = x* x/(0.56-x)

for solving the equation

x=

at equilibrium

concentration of H3O+ = 9.46 x10^-6

[H3O+] = 9.46 x10^-6M

b)

Concentration of Pyridine = C5H5N = 0.558M

Kb of Pyridine = 1.7 x10^-9

Kb = 1.7 x10^-9

                                     C5H5N   + H2O ---------------------- C5H5NH+    +     OH-

Initial                              0.558                                            0                        0

change                             -x                                               +x                     +x

equilibrium                     0.558 -x                                         +x                    +x

kb = [ C5H5NH+] [OH-] / [ C5H5N]

1.7 x10^-9 = x* x/(0.558-x)

for solving the equation

x= 3.08 x10^-5

at equilibrium

[OH-] = 3.08 x10^-5

but [H3O+] [OH-] = Kw               where Kw = ionic prouduct of water = 1.0 x 10^-14

[H3O+] = Kw / [OH-] = 1.0 x10^-14 / 3.08 x10^-5 = 3.25 x10^-10

[H3O+] = 3.25 x10^-10M

2)

a) Concentration of acetyl salicylic acid = HC9H7O4 = 0.445M

Ka of HC9H7O4 = 3.0 x10^-4   

                        HC9H7O4   +   H2O --------------------- C9H7O4-    + H3O+

initial                   0.445                                                0                 0

change                  -x                                                   +x                 +x

equilibrium           0.445-x                                             +x                 +x

Ka = [ C9H7O4-] [H3O+] / [HC9H7O4]

3.0 x10^-4 = x* x/(0 445-x)

for solving the equation

x= 0.0114

at equilibrium

[H3O+] = 0.0114M

-log{H3O+] = -log( 0.0114)

PH = 1.94

but , PH + POH = 14

POH = 14 - PH

POH = 14 - 1.94

POH = 12.06

b) concentration of Aniline = C6H5NH2 = 0.517M

kb of anilinie = 3.8 x 10^-10

                               C6H5NH2   + H2O --------------- C6H5NH3+ + OH-

Initial                        0.517                                         0                 0

change                     -x                                             +x                 +x

equilibrium               0.517 -x                                       +x                +x

Kb = [ C6H5Nh3+] [OH-]/[C6H5NH2]

3.8 x10^-10 = x*x/( 0.517-x)

for solving the equation

x= 1.40 x10^-5

at equilibrium

[OH-] = 1.40x10^-5M

-log[OH-] = -log(1.40x10^-5)

POH = 4.85

PH = 14 - 4.85

PH = 9.15