A chemist titrates of a pyridine solution with solution at . Calculate the pH at equivalence. The of pyridine is . Round your answer to decimal places.

pH = 2.96

Note: The question is incomplete. the complete question is given below:

A chemist titrates 80.0mL of a 0.3184M pyridine C5H5N solution with 0.5397M HBr solution at 25°C . Calculate the pH at equivalence. The pKb of pyridine is 8.77 . Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HBr solution added.

Explanation:

Equation of reaction: C₅H₅NH + HBr > C₅H₅NH⁺ + Br⁻

pKb = -log Kb

8.77= -log Kb

Kb = 1.698 * 10⁻⁹

Volume, V of HBR added to attain equivalence point is obtained:

millimoles of C₅H₅N reacting = 0.3184 M * 80.0 mL  = 25.472 mmol

moles of C₅H₅N = moles of HBr

25.472 mmol = 0.5397 M * V

V = 47.2 mL

25.472 millimoles of both  C₅H₅N and HBr to form C₅H₅NH⁺ and H₂O

C₅H₅NH⁺ formed = 25.472 mmol

Volume of Solution = (80 + 47.2) mL  = 127.2 mL

Concentration of C₅H₅NH⁺, C = 25.472 mmol/127.2 mL = 0.200 M

Ka of C₅H₅NH⁺ = Kw/Kb

Ka = 1.0 * 10⁻¹⁴ /1.698 * 10⁻⁹ = 5.889 * 10⁻⁹

Hydrolysis of C₅H₅NH⁺: C₅H₅NH⁺     +   H₂O >  C₅H₅N  +   H⁺

Initial concentration:          0.200                               0         0

Change in concentration: 0.200 - x                           x         x

Ka = [H⁺][C₅H₅N]/[C₅H₅NH⁺]

Ka = x * x/(c-x)

Assuming x is negligible compared to, the expression above simplifies to

Ka = x²/(c)

x = √(Ka * c)

x = √(5.889 * 10⁻⁶ * 0.200)

x = 1.085 * 10⁻³

Since x = [H+]

[H+] =

Using pH = -log [H+]

pH = -log (1.085 * 10⁻³)

pH = 2.96

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