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!  

answer: e.   sour taste

explanation: acids are those substances which dissociates in water to give ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

bases are those substances which dissociates in water to give ions or accepts ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.