Why does NaOH catalyze H2O2 decomposition? I had a lab practice in which I had to test the effectiveness of different catalysts for the decomposition of $\ce{H2O2}$. Among them, there was: $\ce{KI(aq)}$, $\ce{FeCl3(aq)}$, $\ce{MnO(s)}$, $\ce{NaOH(aq)}$, $\ce{HCl(aq)}$ and $\ce{I2(s)}$.

In different test tubes, I added some $\ce{H2O2}$ and some catalyst. After a while, I added $\ce{NaOH}$ as well. I observed an increase in the effervescence for $\ce{KI}$, $\ce{FeCl3}$, $\ce{I2}$ and $\ce{MnO}$. However, when the mixture was only $\ce{H2O2}$ + $\ce{NaOH}$, nothing happened. I've made some research, but can't find an explanation for this...

I believe $\ce{NaOH}$ increases the rate of the reaction of the other catalysts, but why?
According to THE HOMOGENEOUS BASE-CATALYZED DECOMPOSITION OF HYDROGEN PEROXIDE J. Phys. Chem., 1961, vol. 65 , pages 304–306:

$\ce{H2O2 + OH- &lt;=&gt; HO2- + H2O}$ 

$\ce{H2O2 + HO2- -&gt;}$ [6-membered ring with 2 "H"s bridging 2 "O-O"s, with another H bonded to one of the "O"s] $\ce{-&gt; H2O + O2 + OH-}$


  It is clear that the rate will go through a maximum
  when the peroxide is 50% ionized, that is
  $[\ce{H2O2}] = [\ce{HO2-}]$.


So OH- catalyzes decomposition of peroxide, but a maximum is reacted when the amount of hydroxide added is half the amount of peroxide.  If more hydroxide than peroxide is added, the catalytic effect is lost.