Question

Calculate the titration of 25.0 ml of 0.110M pyridine C5H5N(aq) with 0.110M HBr before the additiom of any Hbr, after the additiion of 12.5 mL of Hbr, after the addition of 23 mL of Hbr after the adddition of 25 mL of Hbr and after the addition of 33 mL of Hbr

Answer 1

Both the concentrations are same and both are the weak Base and Acid.

0.110 M of pyridine (base)

P^OH = -log₁₀ [OH^-]

P^OH = -log₁₀ [0.11]

P^OH = -log₁₀ [11x10^-2]

P^OH =0.95

For the 12.5 ml of HBr:

MV = M1v1 - M2V2 V= Total volume of acid and base

M1V1 is for the Pyridine and M2V2 is for HBr

M (25+12.5) = 0.11x25 - 0.11x12.5

Mx37.5 = 2.75-1.375

M=1.375/37.5

M= 0.036 M of basic in nature of resultant solution.

For the 23 ml of HBr:

Both the concentrations are same but the pyridine volume is more then resultant solution is Basic in nature.

MV = M1v1 - M2V2 V= Total volume of acid and base

M1V1 is for the Pyridine and M2V2 is for HBr

M (25+23) = 0.11x25 - 0.11x23

Mx48 = 2.75-2.53

M=0.22/48

M= 0.0045 M of basic in nature of resultant solution.

For the 25 ml of HBr:

Both the volumes and the concentrations are same, Hence the solution resultant concentration is Zero.

Then the solution is Neutral and its PH is 7.

For the 33 ml of HBr:

Both the concentrations are same but the HBr volume is more then resultant solution is Acidic in nature.

MV = M1v1 - M2V2 V= Total volume of acid and base

M1V1 is for the HBr and M2V2 is for Pyridine

M (25+33) = 0.11x33 - 0.11x25

Mx58 = 3.63-2.75

M=0.88/58

M= 0.015 M of Acidic in nature of resultant solution.