HW15.101 Regarding OH

[Lizzie Zhang 2C]

The reaction in this problem has three steps:
Step 1: ClO-+H2O--->HClO+OH- (Fast equilibrium)
Step 2: HClO+I---->HIO+Cl- (Slow step)
Step 3: HIO+OH- --->IO-+H2O (Fast equilibrium)

The rate law proposed contains OH-. Isn't OH- an intermediate? Since it's produced and then consumed in a later step, why it can show up in the rate law? Also, how should I know that H2O is the solvent and is always present in excess quantity?

Thanks!

[Srbui Azarapetian 2C]

Typically, the rate law would come from the slow step so, R=k2 [HOCL][I-]. However, HOCL is an intermediate, so it needs to be expressed in terms of other products and reactants of the other fast reactions. Using the equation ClO-+H2O--->HClO+OH-, K = [HOCL][OH]/[OCL]. From this equilibrium equation, [HOCL] = [OCL]/ [OH]. Plugging back into the original rate law, R= [OCL] [I-]/ [OH].
Also, water is usually always the solvent and present in greater quantity/excess.

[Lizzie Zhang 2C]

But my question is that if OH- is formed and then consumed in a later step, why wouldn't we count OH- as an intermediate?