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If $\pu{30.0 ml}$ of $\pu{0.20 M}$ ethanol solution reacts with $\pu{1.80 g}$ potassium permanganate, what will be the final $\mathrm{pH}$ of the solution? Assume a complete reaction and that only the products of the reaction contribute to the final $\mathrm{pH}.$

I am not sure what is the correct reaction, so I cannot solve for $\mathrm{pH}.$ Is it

$$\ce{2 KMnO4 + 3 C2H5OH -> 2 MnO2 + 3 CH3CHO + 2KOH + 2H2O}?\tag{R1}$$

But $\ce{KMnO4}$ can also oxidize $\ce{EtOH}$ to a carboxylic acid:

$$\ce{CH3CH2OH + KMnO4 -> CH3COOH + H2O}.\tag{R2}$$

Should I compute $\mathrm{pH}$ based on the $\ce{CH3COOH}?$

$$\ce{CH3COOH(aq) -> CH3COO- + H+}?\tag{R3}$$

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  • $\begingroup$ Rather potassium acetate and hydroxide, as stoichiometry says. $\endgroup$ – Poutnik Feb 8 at 9:11
  • $\begingroup$ I would assume complete oxidation of ethanol to acetic acid. There appears to be enough permanganate to do that and even more. Technically, the medium should be acidified, but the author probably suggested complete oxidation as is. $\endgroup$ – andselisk Feb 8 at 11:38
  • $\begingroup$ If I assume complete oxidation, then I can compute pH from the ionization of acetic acid using Ka and ICE table right? But the answer I am getting is not in the choices. ( 13.6, 13.4, 0.574, 0.421) $\endgroup$ – Gelo Feb 8 at 14:13
  • $\begingroup$ Before doing any ICE calculation, balance the reaction equation first. $\endgroup$ – Poutnik Feb 9 at 8:15

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