Calculate the equilibrium constant for the weak base C5H5N, if a solution of the base with an initial concentration of 5.74×10-3 M has a [C5H5NH+] of 0.00000283 M (make an approximate calculation assuming that initial concentration is equal to the equilibrium concentration).

C5H5N + H2O = C5H5NH+ + OH-

See your HAc problem.

To calculate the equilibrium constant (K) for the reaction, you need to use the concentrations of the products and reactants at equilibrium. In this case, the equation is:

C5H5N + H2O = C5H5NH+ + OH-

Since the given problem assumes that the initial concentration is equal to the equilibrium concentration, you can assume that the concentration of C5H5N at equilibrium is 5.74×10-3 M.

You're also given that the concentration of C5H5NH+ at equilibrium is 0.00000283 M.

Since OH- is the conjugate base of water (H2O), the concentration of OH- can be calculated using the concentration of C5H5N in the given equation.

The concentration of OH- at equilibrium would be equal to the concentration of C5H5N at equilibrium, which is 5.74×10-3 M.

Now, you have the following concentrations at equilibrium:

[C5H5N] = 5.74×10-3 M
[C5H5NH+] = 0.00000283 M
[OH-] = 5.74×10-3 M

To calculate the equilibrium constant (K), you can use the formula:

K = ([C5H5NH+][OH-]) / [C5H5N]

Substituting the given concentrations into the formula:

K = (0.00000283 × 5.74×10-3) / 5.74×10-3

The concentration of C5H5N in the numerator cancels out with the concentration of C5H5N in the denominator, resulting in:

K = 0.00000283

Therefore, the equilibrium constant (K) for the reaction is approximately 0.00000283.