Question

a. If the pH = 2.42, what is the [H₃O⁺]? Remember that [H₃O⁺] = antilog of (- pH).
[H₃O⁺] = = x

b. 1.80 M solution of CH₃NH₃⁺ has a pH = 1.554. What is the equilibrium constant for this species in water?

c. What is the pH of a 0.748 M solution of C₆H₅COOH if has a percent ionization = 0.918%? You don't need a K value here.

Answer 1

a. [H3O+] = 10^-pH

         = 10^-2.42

       = 0.0038 M

b. CH3NH3+ is acidic, pH = 7-1/2(pkb+logC)

   1.554 = 7-1/2(x+log1.8)

   pkb of CH3NH2 = 10.64

   pka of CH3-NH3+ = 14-10.64 = 3.36

    ka of CH3-NH3+ = 10^-3.36 = 4.36*10^-4

C. [H3O+] = CX

   X = degree of dissociation = 0.918/100 = 0.00918

   [H3O+] = 0.748*0.00918 = 6.87*10^-3 M

   pH = -log(6.87*10^-3) = 2.163