HIO₃ behaves as acid in water: HIO₃(aq)$\leftrightarrows$ IO₃^-(aq)+ H⁺(aq),with K_c = 0.17 at 25 $\circ$ C.What is the H⁺ concentration in a solution that is initially 0.50 M HIO₃?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the dissociation of HIO₃ is Kc = [H⁺][IO₃⁻]/[HIO₃]. Given Kc = 0.17 and initial [HIO₃] = 0.50 M, we can set up the equation 0.17 = x²/(0.50 - x), where x is the concentration of H⁺ at equilibrium. Solving for x gives approximately 0.22 M.

HIO3 Behaves as Acid in Water: HIO3(aq) ⇆\leftrightarrows⇆ IO3-(aq)+ H+(aq),with Kc = 0

HIO₃ Behaves as Acid in Water: HIO₃(aq) $\leftrightarrows$ IO₃^-(aq)+ H⁺(aq),with K_c = 0