I have performed an experiment where I added an excess of $\ce{Ca(OH)2}$ base to a solution consisting of 5 mL of 30 % hydrogen peroxide (buffered at pH 5) and a very small amount of sulfuric acid (such that the pH of the original solution of 30 % $\ce{H2O2}$ and acid was 1.7).

I am having trouble understanding what reactions maybe occurring between these three substances to give a final pH of 10.5.

I know that $\ce{Ca(OH)2}$ and $\ce{H2O2}$ form $\ce{CaO2}$ when reacted, but shouldn't the final pH be equal to that of $\ce{Ca(OH)2}$ considering it is in excess? What reactions could be occurring here and why are they not allowing the pH to reach that of $\ce{Ca(OH)2}$?

Any help at all would be very much appreciated. Thank you in advance.

  • 1
    $\begingroup$ There is no CaO2. $\endgroup$ – Ivan Neretin Sep 1 '19 at 3:48
  • 2
    $\begingroup$ @Ivan Neretin en.wikipedia.org/wiki/Calcium_peroxide Calcium peroxide is produced by combining calcium salts and hydrogen peroxide: Ca(OH)2 + H2O2 → CaO2 + 2 H2O The octahydrate precipitates upon the reaction of calcium hydroxide with dilute hydrogen peroxide. Upon heating it dehydrates. $\endgroup$ – Poutnik Sep 1 '19 at 4:18
  • 1
    $\begingroup$ All the same, I don't believe it will form in the said conditions. $\endgroup$ – Ivan Neretin Sep 1 '19 at 4:21
  • 2
    $\begingroup$ It is not the same as "there is no CaO2". $\endgroup$ – Poutnik Sep 1 '19 at 4:24
  • $\begingroup$ Let me add that, if pH is 10.5, it's clearly not "neutralized pH". As for the final question, you should read about buffers. $\endgroup$ – The_Vinz Sep 1 '19 at 4:49

in your question formulation, you have forgotten to take into account $\ce{H2O2}$ is a weak acid.

The title should rather be: Neutralisation between calcium hydroxide and 30% hydrogen peroxide"

Unless $\ce{Ca(OH)2}$ was in excess over $\ce{H2O2}$ - and it was said it was not - $\mathrm{pH}$ would be always significantly lower than pH of the hydroxide.

$$\mathrm{pH}=\mathrm{p}K_ \mathrm{a,\ce{H2O2}} + \log \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$$

where $\mathrm{p}K_ \mathrm{a,\ce{H2O2}}=11.75$$ by Wikipedia, but see the links below.

If we consider reaction

$$\ce{Ca(OH)2 + H2O2 -> H2O + Ca(OH)(HO2)}$$

we need to neutralize 50% of $\ce{H2O2}$ to reach $\mathrm{pH}=\mathrm{p}K_ \mathrm{a,\ce{H2O2}}$

The hydroxide forms from $\ce{H2O2}$ the $\mathrm{pH}$ buffer solution of a weak acid and it's salt.

$$\begin{align} \ce{Ca(OH)2 &<=>> CaOH+ + OH-}\\ \ce{CaOH+ &<=>> Ca^2+ + OH- }\\ \ce{H2O2 &<<=> H+ + HO2-}\\ \ce{H+ + OH- &<=>> H2O}\\ \end{align}$$

$\ce{Ca(OH)2}$: $\mathrm{p}K_\mathrm{b1} =1.37$, $\mathrm{p}K_\mathrm{b2} =2.43$ ( Wikipedia )

Additionally, $\ce{HO2-}$ is partially eliminated by precipitation, therefore ratio $ \frac{[\ce{HO2-}]}{[\ce{H2O2}]}$ is kept low and so does $\mathrm{pH}$.

$$\ce{CaOH+ + HO2- + 7 H2O <=>> CaO2 \cdot 8 H2O v}$$

Note also the hydrogen peroxide is weakly acidic even without addition of sulphuric acid and that it's $ \mathrm{p}K_ \mathrm{a}$ depends on $\ce{H2O2}$ concentration.

H2O2 pH-and-Ionization-Constant

The solubility constant of calcium peroxide octahydrate in relation to temperature; its influence on radiolysis in cement-based materials

  • $\begingroup$ Does the pKa in the pH equation refer to H2O2 or Ca(OH)2? Thank you. $\endgroup$ – BigDog12 Sep 9 '19 at 5:12
  • $\begingroup$ H2O2. for Ca(OH)2, there would calcium compounds instead. $\endgroup$ – Poutnik Sep 9 '19 at 5:14
  • $\begingroup$ Okay. Thank you very much for all of your help. $\endgroup$ – BigDog12 Sep 9 '19 at 8:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.