Answer:
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.
Step-by-step 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