Find the initial concentration of the weak

acid or base in an aqueous solution of pyridine
(C5H5N) with a pH of 8.65. Kb = 1.80×10e-9
.
Answer in units of mol/L

drbob where the hell is your HClO problem

To find the initial concentration of the weak acid or base in the aqueous solution of pyridine (C5H5N) with a pH of 8.65, we can use the relationship between pH and pOH, as well as the equilibrium constant for the base, Kb.

The pOH can be calculated using the formula:
pOH = 14 - pH

Given that the pH is 8.65, we can find the pOH:
pOH = 14 - 8.65
pOH = 5.35

Next, we can use the relationship between pOH, concentration of hydroxide ions (OH-), and Kb to find the concentration of OH-:
pOH = -log[OH-]
5.35 = -log[OH-]

Rearranging the equation, we can find [OH-]:
[OH-] = 10^(-pOH)
[OH-] = 10^(-5.35)

Now that we have the concentration of OH-, we can use the Kb value given for pyridine to calculate the initial concentration of the weak base.

The Kb expression for the reaction between pyridine and water can be written as:
Kb = [NH4+][OH-] / [C5H5N]

Since pyridine is a weak base, we can assume that the concentration of NH4+ ions is negligible compared to the initial concentration of pyridine (C5H5N). Therefore, we can simplify the Kb expression to:
Kb = [OH-] * [C5H5N]

Now, substitute the values of [OH-] and Kb in the equation and solve for [C5H5N]:
1.80×10^(-9) = (10^(-5.35)) * [C5H5N]

To isolate [C5H5N], divide both sides of the equation by 10^(-5.35):
[C5H5N] = (1.80×10^(-9)) / (10^(-5.35))
[C5H5N] = 1.26×10^(-15)

Therefore, the initial concentration of the weak base pyridine in the aqueous solution is 1.26×10^(-15) mol/L.

Do this the same way as the HClO problem.