I recognize that I need to identify the equipment that has the conjugate base of a weak acid.

You are watching: Is ch3oh an acid or base

Knowing this, I immediately eliminated $\ceBaCl2$ ($\ceHCl$), $\ceK2SO4$ ($\ceH2SO4$), and $\ce(NH4)2CrO4$ ($\ceH2CrO4$ — i beg your pardon I"ve heard is a strong acid).

This left $\ceCH3OH$ and also $\ceNa2CO3$.

The correct answer is $\ceNa2CO3$ because $\ceCO3^2-$ is the conjugate basic of $\ceH2CO3$.

However, why have the right to the last answer not be $\ceCH3OH$, especially considering the $\ceOH$ is basic? Also, can $\ceOH$ be thought about the conjugate basic to a weak acid?

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edited Jul 7 "17 at 5:38

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Unfortunately, bsci-ch.org is no as straightforward as going

Oh, i spotted $\ceOH$, it have to be a basic compound!

I would certainly argue the the vast bulk of every compounds one could write v $\ceOH$ inside are either neutral or acidic and only a minority is basic.

Take, for example, sulphuric acid. As Ivan correctly pointed out in a comment, quite than writing the traditional $\ceH2SO4$, you could also write it as $\ceO2S(OH)2$ which lays an ext emphasis top top the molecular structure (to hydroxy groups and two oxy teams all bound to sulphur). We all know that $\ceH2SO4$ is acidic, not basic, even though it has actually $\ceOH$ groups.

Or take alcohols such as methanol ($\ceH3C-OH$) or ethanol ($\ceH3C-CH2-OH$). These likewise feature $\ceOH$ groups, they room even commonly written the way, but are around as acidic or an easy as water ($\ceHOH$) is.

So the question may be why did you obtain taught the $\ceOH$ is basic? fine strictly speaking, the is not the $\ceOH$ hydroxy team that is basic but the $\ceOH-$ hydroxide ion. Just those compounds that dissociate into some cation and also a hydroxide ion have the right to be considered basic. For that to happen, ‘some cation’ typically needs to be metallic, at the very least at your level. Therefore recognise a metal external inspection to $\ceOH-$ and you have to most likely be best calling the compound basic.

Some last notes:

You should always check the directly corresponding conjugate acid. So once analysing $\ceK2SO4, Na2CO3$ and also $\ce(NH4)2CrO4$, trace them ago to $\ceHSO4- , HCO3-$ and also $\ceHCrO4-$, respectively. That still pipeline us v two rather strong acids yet the third ($\ceHCO3-$) is also weaker.

The conjugate mountain to methanol $\ceH3C-OH$ would be the methyloxonium ion $\ceH3C-OH2+$, i m sorry is a solid acid. This goes because that pretty much all necessary hydroxy groups.

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The $\mathrmpK_\mathrma$ of water — hydroxide’s conjugate acid — is $15.7$, so yes, that is a really weak acid.