Calculate the pH of each of the following solutions. a. 0.100 M propanoic acid (HC3H5O2, Ka 1.3 105 ) b. 0.100 M sodium propanoate (NaC3H5O2) c. pure H2O d. a mixture containing 0.100 M HC3H5O2 and 0.100 M NaC3H5O2

a) pH = 2.95

b) pH = 8.94

c) pH = 7.00

d) pH = 4.89

Explanation:

a) The reaction is:

               CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)(1)

At equlibrium:              0.1 - x                                     x                       x

The dissociation constant of equation (1), Ka, is:

  (2)

Solving equation (2) for x, we have:

x = 0.00113 = [H₃O⁺] = [CH₃CH₂COO⁻]      

Hence, the pH is:

b) The reaction is:  

            CH₃CH₂COO⁻(aq) + H₂O(l)  ⇄ CH₃CH₂COOH(aq) + OH⁻(aq)

At equlibrium:    0.1 - x                                        x                         x  

The dissociation constant of the above equation, Kb, is:  

Also, we have that:

   (3)

Solving equation (3) for x, we have:

x = 8.77x10⁻⁶ = [OH⁻] = [CH₃CH₂COOH]

Hence, the pH is:

c) Pure water:

H₂O ⇄ H⁺ + OH⁻

Since is pure water: [H⁺] = [OH⁻]  

Therefore, the pH is:

d) The reaction is:

          CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)  

At equlibrium:           0.1                                     0.1                        x

The pH is:

I hope it helps you!  

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