The S^2- ion is a relatively strong base with an equilibrium constant of 7.7x10^-1.

What is the Ka value for HS- ?

I got the equation HS^- <-> H^+ + S^-2. Am I able to find [S^-2] with the Keq?

I think Ka = Kw/Kb. I'm not sure what you mean by S%2= ion has eq const of 0.77. Is that Kb?

S^-2 ion has equilibrium constant of 7.7x10^-1

So it's just (1.00x10^-14)/(0.77)=1.3x10^14 = Ka HS^- ?

If the equilibrium constant is the same as Kb, yes. But I'm not sure that it is. I've never seen an ION shown as an equilibrium constant.

Alright, thanks! It's the correct answer on the answer key.

To find the Ka value for HS-, you need to use the equilibrium constant expression for the dissociation of HS- into H+ and S^-2:

Ka = [H+][S^-2]/[HS^-]

Before proceeding, it is important to clarify a few things. In the given equation, HS^- refers to hydrogen sulfide, not the S^2- ion. The equilibrium constant you provided, 7.7x10^-1, is actually Kc (equilibrium constant in terms of concentrations) rather than Keq (equilibrium constant in terms of activities).

To find the Ka value, you need the concentration of [H+], [S^-2], and [HS^-]. However, without knowing the initial concentrations or any changes in equilibrium concentrations, it is not possible to directly determine the individual concentrations of [H+], [S^-2], and [HS^-].

The given equilibrium constant value, 7.7x10^-1, does not directly provide enough information to solve for the Ka value of HS-. It only indicates the relative strength of the base compared to its reactants.

To determine the exact value of Ka for HS^-, you will need additional information such as concentrations or the equilibrium concentrations of the species involved in the reaction.