Question

A 50.00-mL solution of 0.100 M ethanolamine (HOC2H5NH2) is titrated with a standardized 0.200 M solution of HCl at 25°C. Enter your numbers to 2 decimal places. Kb = 3.2 x 10-5

Answer with work please :)

1. What is the pH of the ethanolamine solution before titrant is added?

2. What is the pH at the half-equivalence point?

3. What is the pH at the equivalence point?

4. What is the pH after 30.00 ml of 0.200 M HCl has been added to 50.00 mL solution of 0.100 M ethanolamine?

Answer 1

basic buffer

1. before titrant is added

pH of weak base = 14 - (1/2(pkb-logC))

        pkb = -logkb

            = - log(3.2*10^-5)

            = 4.5

    pH = 14 - (1/2(4.5-log0.1))

       = 11.25

2. at the half-equivalence point of basic buffer

    pOH = pKb

pH = 14 - pOH

     = 14 - 4.5

     = 9.5

3. at the equivalence point

no of mole of ethanolamine (HOC2H5NH2) = no of mole of HCl

no of mole of ethanolamine (HOC2H5NH2) = 50*0.1/1000 = 0.005 mole

volume of HCl must take = n/M = 0.005 / 0.2 = 0.025 L = 25 ml

pH of weakbase strongacid salt = 7 - 1/2(pkb+logC)

        c = concentration of salt = 0.005*1000/(25+50) = 0.067 M

pH = 7 - 1/2(4.5+log0.067)

   = 5.34

4. no of mole of excess HCl added = (30*0.2 - 25*0.2)/1000

                = 0.001 mole

   concentration of excess HCl = n/V

                 = 0.001 / 0.075

                 = 0.013 M

    pH = -log(H+) = -log0.013 = 1.89