the equilibrium constant for this reversible reaction is 3.6*10^-7

OCl^-(aq)+ H₂O gives u HOCl(aq) + OH^-(aq)

what is K_a for HOCl?

We are given the equilibrium constant for the reaction:

OCl⁻(aq) + H₂O ⇌ HOCl(aq) + OH⁻(aq)
K = 3.6 × 10⁻⁷

We need to determine the acid dissociation constant (Ka) for HOCl.

Step 1: Identify the relationship between Ka and Kb
The reaction given represents the hydrolysis of the hypochlorite ion (OCl⁻), which is the conjugate base of hypochlorous acid (HOCl). The base dissociation constant (Kb) for OCl⁻ is related to the acid dissociation constant (Ka) of HOCl through the equation:

Ka × Kb = Kw

where Kw is the ionization constant of water, which is:

Kw = 1.0 × 10⁻¹⁴ at 25°C.

Since the given equilibrium constant (K) represents Kb for OCl⁻, we substitute:

Ka × (3.6 × 10⁻⁷) = 1.0 × 10⁻¹⁴

Step 2: Solve for Ka
Ka = (1.0 × 10⁻¹⁴) / (3.6 × 10⁻⁷)
Ka = 2.78 × 10⁻⁸

Final Answer:
The acid dissociation constant (Ka) for HOCl is 2.78 × 10⁻⁸.