Question

1. a. Write the equation for the dissociation of the weak base methylamine, CH3NH2.

b. What is the pH of a 1.556M solution of methylamine?

c. What is the pH of a buffer made by mixing 1.556 mol methylamine with 1.556 mol methylammonium chloride (CH3NH3+).

d. What is the pH of this buffer after the addition of 0.0056mol HCl?

e. What is the pH of this buffer after the addition of 0.0056 mol NaOH?

Answer 1

a. CH3-NH2 (aq) + H2O(l) ---> CH3-NH3+(aq) + OH^-(aq)

b. pH of CH3-NH2 = 14 - 1/2(pkb-logC)

            pkb of ch3-nh2 = 3.38

         C = concentration of CH3-NH2 = 1.556 M

    pH = 14 - (1/2(3.38-log1.556)) = 12.4

c. pH of buffer = pka + log(base/salt)

       pka of CH3-NH2 = 14-3.38 = 10.62

   pH = 10.62+log(1.556/1.556)

      = 10.62

d. after the addition of 0.0056mol HCl

   pH of buffer = pka + log((base+HCl/salt-HCl)

       pka of CH3-NH2 = 14-3.38 = 10.62

   pH = 10.62+log((1.556+0.0056)/(1.556-0.0056))

      = 10.623

e. after the addition of 0.0056 mol NaOH

   pH of buffer = pka + log((base-NaOH/salt+NaOH)

       pka of CH3-NH2 = 14-3.38 = 10.62

   pH = 10.62+log((1.556-0.0056)/(1.556+0.0056))

      = 10.617