(1) ClO-(aq) + H2O(l) HClO(aq) + OH -(aq) [fast] (2) I -(aq) + HClO(aq) HIO(aq) + Cl -(aq) [slow] (3) OH -(aq) + HIO(aq) H2O(l) + IO-(aq) [fast] (a) What is the overall equation? (Type your answer using the format [NH4]+ for NH4+. Use the lowest possible coefficients. Enter 0 if necessary. Do not leave any box blank.)

ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + IO⁻(aq)

Explanation:

The reactions are:

(1) ClO⁻(aq) + H₂O(l) → HClO(aq) + OH⁻(aq)    [fast]

(2) I⁻(aq) + HClO(aq) →  HIO(aq) + Cl⁻(aq)  [slow]

(3) OH⁻(aq) + HIO(aq) → H₂O(l) + IO⁻(aq)  [fast]

By adding up the 3 equations we get:

ClO⁻(aq) + H₂O(l) + I⁻(aq) + HClO(aq) + OH⁻(aq) + HIO(aq)  →  HClO(aq) + OH⁻(aq) + HIO(aq) + Cl⁻(aq) + H₂O(l) + IO⁻(aq)

And by canceling common terms on both sides, we can get the overall equation:

ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + IO⁻(aq)  

Therefore, the overall equation is ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + IO⁻(aq).

I hope it helps you!