If K = 1.8 x 10 for HC2H3O (aq) <=> H+(aq)+C2H3O2-(aq), and 100 mL of aqueous solution has [HC2H3O2] = 0.50 M, [H+] = 0.090 M, and [C2H3O2-] 0.0010 M, is the system at equilibrium?

I think you omitted a -5 from the 10^-5 for Ka

HC2H3O2 ==> H^+ + C2H3O2^-

Ka = (H^+)(C2H3O2^-)/(HC2H3O2)
Plug in the values given in the problem and see if you get Ka. If you do then the system is at equilibrium. If not, the system is not at equilibrium.

To determine whether the system is at equilibrium, we need to compare the calculated value of K with the given concentrations of the reactants and products.

First, let's write down the expression of the equilibrium constant (K) for the given reaction:

K = [H+][C2H3O2-] / [HC2H3O2]

Now, let's substitute the given concentrations into the equation and calculate the value of K:

K = (0.090)(0.0010) / (0.50)

K = 0.000090 / 0.50

K = 0.00018

Now, compare the calculated value of K with the given value of K, which is 1.8 x 10^(-10).

Since the calculated value of K (0.00018) is higher than the given value of K (1.8 x 10^(-10)), we can conclude that the system is not at equilibrium.

In other words, the concentrations of the reactants and products do not correspond to the equilibrium state for this reaction.