Reaction Mechanism with a Rate Determining Step

Just a quick question on how to deal with multiple intermediaries in a proposed reaction mechanism.


Here is the overall reaction:


$$\ce{I-(aq) + OCl-(aq) -> IO-(aq) + Cl-(aq)}$$


Here is the proposed mechanism:


$$
\begin{align}
\ce{OCl- + H2O &<=>[$k\_1$][$k\_{-1}$] HOCl + OH-} & &\text{(Fast equilibrium)} \\ 
\ce{I- + HOCl &->[$k\_2$] HOI + Cl-} & &\text{(Slow/RDS, non-reversible)} \\
\ce{HOI + OH- &->[$k\_1$] H2O + IO-} & &\text{(Fast)}
\end{align}
$$


A few things to note, the question does indeed label both the first reaction going forwards and the third reaction going forwards as $k\_1$. It also mentions that we should treat the concentration of water as being constant.


Here is my question:
I know how to write the rate law for the Rate Determining Step as $k\_2[\ce{I-}][\ce{HOCl}]$ and then I know I need to use the equilibrium expression in the first reaction to isolate concentration of $[\ce{HOCl}]$ as


$$[\ce{HOCl}] = \frac{k\_1[\ce{OCl-}]}{k\_{-1}[\ce{OH-}]}$$


but then I am confused about how to isolate the second intermediary $(\ce{OH-})$ so I can get the proper rate law. Should I use the third reaction?


