Question

30.0 mL solution of 0.010 M piperidine (the active ingredient in hot pepper, the protonated form is a monoprotic weak acid with pKa= 11.123) was titrated with 0.050 M HCl.   a) Find Ve (the volume of HCl at equivalent point); b) Calculate the pH at Va (the volume of HCl added) =0, ¼ Ve, Ve, and 1.1 Ve; c) Using the information from (b), sketch the titration curve for this titration. Please mark the “landmark” points on this curve.

Answer 1

a) when Volume of HCl = 0mL

There will be only piperidine (B) in the solution

It will hydrolyze as

            B + H2O ---> BH⁺ + OH-

Initial    0.01                  0           0

Change -x                   +x         +x

Equilib   0.01-x              x           x

Kb = Kw / Ka

pKa = 11.123   pKb = 14- pKa = 14 - 11.123 = 2.877

Kb = antilog (-2.877) = 1.33 X 10^-3

Kb= [OH-] [ BH⁺ ]/ [B] = x² / (0.01-x) = 1.33 X 10^-3

x² = 0.0000133 - 0.00133x

x² - 0.0000133 + 0.00133x = 0

x = 0.03 M = [OH-]

pOH = -log[OH-] = 1.52

pH = 14- 1.52 = 12.48

b) Ve, when the moles of HCl = moles of piperidine

Moles of piperidine = Molarity X volume = 0.01 X 30 /1000 = 0.0003 moles

so moles of HCl at point of equivalence = 0.0003 moles

Ve = Moles / molarity = 0.0003 / 0.05 = 0.006 L = 6mL

Ve/4 = 1.5 mL of HCl

The moles of Hcl = 0.05 X 1.5 = 0.000075 moles

It will react with the same number of moles of piperidine to give same number of moles of salt of piperidine

The moles of piperidine left = 0.0003 - 0.000075 = 0.000225 moles

it will form a buffer,

pOH = pKb + log [salt] / [base]

pOH = 2.877 + log [0.000075 / 0.000225] = 2.4

pH = 14 - pOH = 14 - 2.4 = 11.6

c) At ve

There will be formation of salt of piperidine which will hydrolze further

Moles of piperidine salt formed = moles of HCl added = 0.0003

[Salt] = Moles / total volume = 8.33 X 10^-6 M

Let salt is B+Cl-

                     B⁺    +        H2O ---> BOH   +    H⁺

Initial conc     8.33 X 10^-6 0 0

change          -x                                       +x              +x

Equilibrium    8.33 X 10^-6 -x                     x                   x

Ka = [H⁺] [ BOH] / [B⁺]

7.53 X 10^-12 = x² / (8.33 X 10^-6 -x)

we can ignore x in denominator as Ka is very low

7.53 X 10^-12 = x² / 8.33 X 10^-6

x = 7.92 X 10^-9 = [H⁺]

As it is very low we cannot ignore the contributtion of H+ by water (10-7)

[H⁺] = 7.92 X 10^-9 + 10^-7

pH = -log [H⁺] = 6.97

d) at 1.1 Ve

the volume of Hcl added = 1.1 X 6 = 6.6 mL

so the volume of Hcl which will not react with piperidine = 0.6 mL

Moles of Hcl unreacted = 0.6 X 0.05 = 0.03 millimoles

[H+] = [HCl] = Moles / Total volume = 0.03 millimoles /36.6mL = 8.197 X 10^-4 M

pH = -log[H+] = 3.09