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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.

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    $\begingroup$ There is no CaO2. $\endgroup$ – Ivan Neretin Sep 1 '19 at 3:48
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    $\begingroup$ @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
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    $\begingroup$ All the same, I don't believe it will form in the said conditions. $\endgroup$ – Ivan Neretin Sep 1 '19 at 4:21
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    $\begingroup$ 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
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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

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  • $\begingroup$ 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

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