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!  

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

so since they all have the same shape? i mean maybe because m& m and reeses are both chocolates