Question

A 30.0 mL sample of .200 M NH3 is titrated with .100M HI. The Kb for NH3 is 1.76X10^-5

A. what is the initial pH of the solution?

B. What is the pH of the solution after the addition of 30.00 mL of HI?

C. What is the pH of the solution at the equivalence point?

D. What is the pH of the solution after the addition of 90.0mL of HI?

Answer 1

pkb = -log(1.76*10^-5) = 4.75

a. initial pH = 14-1/2(pkb-logc)

            = 14-1/2(4.75-log0.2)

          = 11.27

b. No of mol ofNH3 = 30/1000*0.2 = 0.006 mol

No of mol of HI = 30/1000*0.1 = 0.003 moL

pH = 14-(pkb_log(salt/base)

    = 14-(4.75+log(0.003/0.003))

    = 9.25

c. at the equivalence point

concentration of salt = 0.006/0.09 = 0.067 M

pH = 7-1/2(pkb+logC)

    = 7-1/2(4.75+log0.067)

    =5.21

d. concentration of excess HI = 30/120*0.1 = 0.025 M

pH = -log0.025 = 1.6