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?

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.

  • $\begingroup$ I would not think a solution as dilute as 3% would be changed much over an "hours" timescale. By all means, it will decompose, I just cannot see how the rate of that happening is anything but negligible. NB: I assume well out of reach from sunlight. $\endgroup$ – Stian Yttervik May 3 '17 at 11:56

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.

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