Question

You have 25.00 mL of a 0.22 M of a weak base, anline C6H5NH2 with a Kb= 3.94 x 10 ^-10.it is to be titrated with 0.20 M HCl. What is the resulting solution after the following additions? a) intial pH b) after 10.05mL HCl have been added. c) 1/2 way to the equivalence point. d) At the equvalence point. e) 5.0mL after the equivalence point f) what indicator would you use to signal the endpoint and why?

Answer 1

millimoles of aniline = 25 x 0.22 = 5.5

kb= 3.94 x 10^-10

pKb = -logKb = -log 3.94 x 10^-10) = 9.40

a) initial pH

pOH = 1/2 [pKb -logC]

         = 1/2 [9.40 -log0.22] = 5.03

pH + pOH = 14

pH = 8.97

b) after the addition of 10.5 mL HCl

millimoles of HCl = 10.5 x0.20 = 2.1

C6H5NH2 + HBr ----------------------> C6H5NH3+

     5.5          2.1                                 0

   3.4         0                                   2.1

pOH = pKb + log [salt / base]

       = 9.40 + log [2.1/3.4]

pOH = 9.19

pH = 4.81

c) 1/2 way to the equivalence point

it is half equivalence point . so

pOH = pKb

pOH = 9.40

pH +pOH =14

pH = 4.60

c) At the equvalence point

volume of acid = 27.5

here salt remains.

salt concentration = 5.5 / 25 + 27.5 = 0.105 M

pH = 7 - 1/2 (pKb + log C)

     = 7 - 1/2 (9.40 + log 0.105)

pH = 2.79

d) after the addition of 32.5 mL HCl

millimoles of HBr = 32.5 x 0.20 = 6.5

strong acid remained in the solution

[H+] = 1.00 / (25+32.5) = 0.0174 M

pH = -log(0.0174)

pH = 1.76