Final answer:
The equilibrium concentration of H+ for the given reaction with Kc = 10.0 is calculated using an ICE table and the assumption that the change in HClO3 concentration is much less than the initial value. Solving the equilibrium expression for the dissociation of HClO3 gives an H+ concentration of 3.16 M at equilibrium.
Step-by-step explanation:
To calculate the equilibrium concentration of H+ for the reaction HClO3(aq) <----> ClO3-(aq) + H+(aq) with equilibrium constant Kc = 10.0, we can set up an ICE (Initial, Change, Equilibrium) table and solve for the equilibrium concentrations. Since the initial concentration of HClO3 is 1.00 M, we let the change in concentration of HClO3 be -x. Consequently, the change in concentration of ClO3- and H+ will be +x. At equilibrium, the concentrations will be:
- [HClO3] = 1.00 - x
- [ClO3-] = x
- [H+] = x
The equilibrium expression is:
Kc = [ClO3-][H+] / [HClO3]
Substituting the known values and solving for x:
10.0 = (x)(x) / (1.00 - x)
Because Kc is relatively large, we can make the assumption that x will be much less than 1.00 and therefore, 1.00 - x is approximately 1.00. This simplifies to:
10.0 = x2
Taking the square root of both sides gives us x = √10.0 = 3.16. Since [H+] = x, the equilibrium concentration of H+ is 3.16 M.