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This question already has an answer here:

Let's talk about the following compounds - water and methanol.

  1. We know that methanol is slightly more acidic than water, because water's ability to donate a proton as an acid is reduced due to extensive hydrogen bonding. The $\mathrm{p}K_\mathrm{a}$ values are approximately $15.5$ and $15.7$, for methanol and water respectively.

  2. Now consider their conjugate bases, i.e. methoxide ion and hydroxide ion. It is very clear that methoxide is a stronger base compared to hydroxide, due to the +I (inductive) effect of the methyl group (electron releasing).

How is this possible? Isn't it contradictory? It is a well known fact that a strong acid has a weak conjugate base, and a weak acid has a strong conjugate base.

This is a consequence of $\mathrm{p}K_\mathrm{a}$ + $\mathrm{p}K_\mathrm{b}$ = $\mathrm{p}K_\mathrm{w}$, where Kw is the ionic product of water, a constant at a given temperature.

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marked as duplicate by M.A.R. ಠ_ಠ, Tyberius, Mithoron, aventurin, Todd Minehardt Mar 28 '18 at 16:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It's not a duplicate, hence the downvote is completely not justified. $\endgroup$ – arya_stark Mar 28 '18 at 5:40
  • $\begingroup$ In "methanol is slightly more acidic than water, because [...]" and "methoxide is a stronger base compared to hydroxide, due to [...]", "should be" would be more appropriate than "is" -- the "should be" only becomes an "is" once you account for the superposition of the various relevant effects in an actual system, including, in particular, the influence of the solvent. $\endgroup$ – duplode Mar 28 '18 at 5:40
  • $\begingroup$ It's not a duplicate! None of the links posted in the comments have mentioned the comparison between methoxide and hydroxide basic strength comparison - so why are you spamming this post? $\endgroup$ – arya_stark Mar 28 '18 at 5:42
  • $\begingroup$ (1) Note that me and Avnish Kabaj have linked to two different questions -- I believe the second one addresses more directly why the contradiction you point out is only apparent. (2) If the contradiction is only apparent, there is no need to make a separate mention of basic strength -- that is why the linked Q&As don't do that. $\endgroup$ – duplode Mar 28 '18 at 5:47
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  1. We know that methanol is slightly more acidic than water, because water's ability to donate a proton as an acid is reduced due to extensive hydrogen bonding. [...]

  2. [...] It is very clear that methoxide is a stronger base compared to hydroxide, due to the +I (inductive) effect of the methyl group (electron releasing).

You are pointing out competing factors that, in abstract, should lead to either methanol or water being more acidic than the other (that is. lead to either hydroxide or methoxide to be more basic than the other, respectively). Once we move from considering properties in abstract to an actual system in which an acid-base equilibrium is present, the competing factors are superimposed, with a non-contradictory result: either methanol will be more acidic (and hydroxide more basic) or water will be more acidic (and methoxide more basic), depending on which factors dominate. (In particular, as pointed out in Why is methanol more acidic than water? and Is methanol really more acidic than water?, the relative acidities depend on the solvent.)

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