Question

consider the titration of 50.00 ml of .100M trimethylene kb+6.25x10 -5 with .100 M HI. Calculate the pH at initial point.

consider the titration of 50.00 ml of .100M trimethylene kb+6.25x10 -5 with .100 M HI. Calculate the pH after addition of 25.00 ml HI

consider the titration of 50.00 ml of .100M trimethylene kb+6.25x10 -5 with .100 M HI. Calculate the pH after addition of 30.00 ml HI

Answer 1

Let,

trimethylene = B-

Reaction is:

B- + H2O ----> BH + OH-

0.1 - X.............X........X

Kb = [BH] [OH-] / [B-]

=> 6.25 x 10^-5 = X^2 / (0.1 - X)

=> X = 2.5 x 10^-3 M = [OH-]

pOH = - log [OH-] = - log [2.5 x 10^-3] = 2.6

pH = 14 - pOH

=> pH = 11.4

pH after addition of 25.00 ml HI

B- + HI ----> BH + I-

Millimoles of HI added = 25 x 0.1 = 2.5

Millimoles of Base = 50 x 0.1 = 5

Millimoles of Base remaining = 5 - 2.5 = 2.5

Millimoles of BH produced = 2.5

It acts as a Basic Buffer.

pOH = pKb + log (2.5 / 2.5)

=> pOH = 4.2 + log 1

=> pOH = 4.2

pH = 14 - 4.2 = 9.8

pH after addition of 30.00 ml HI

B- + HI ----> BH + I-

Millimoles of HI added = 30 x 0.1 = 3

Millimoles of Base = 50 x 0.1 = 5

Millimoles of Base remaining = 5 - 3 = 2

Millimoles of BH produced = 3

It acts as a Basic Buffer.

pOH = pKb + log (3 / 2)

=> pOH = 4.2 + 0.176

=> pOH = 4.376

pH = 14 - 4.38 = 9.62