I have read online that $\ce{H2O2}$ decomposes if exposed to sunlight and air. If this is the case then will the concentration of 3% w/v of $\ce{H2O2}$ be lowered if it is left out in the open at around 25-degree celsius for a few hours?
Background:
The question arises from a titration experiment that I conducted where I had to verify the concentration of $\ce{H2O2}$ to be 3% w/v. But the titration found the concentration to be lower and other groups in the class found the same.
Hydrogen peroxide is somewhat unstable even when not exposed to air and sunlight. It decomposes to water and oxygen gas over time.
According to this Wikipedia article:
Hydrogen peroxide is thermodynamically unstable and decomposes to form water and oxygen with a $\mathrm{\Delta H}$ of $\mathrm{-98.2}~\pu{kJ/mol}$ and a $\mathrm{\Delta S}$ of $\mathrm{70.5~\pu{kJ/mol}\ce.\pu K}$:
$$\ce{2H2O2 -> 2H2O + O2}$$
The rate of decomposition increases with rising temperature, concentration and pH, with cool, dilute, acidic solutions showing the best stability.
So, not knowing the exact conditions during the 3-week period or other factors, it's not possible to say for certain whether the $\ce{H2O2}$ degraded or whether it just was a little low to begin with. But the potential for degradation over that period of time is certainly there.
