Why is water more acidic than alkanols? Why is $\ce{H-OH}$ more acidic than $\ce{CH3CH2-OH}$? A common answer to this question is because of the $+I$-effect of $\ce{CH3}$ the negative charge at $\ce{O}$ increases making it less acidic. But if we compare electronegativity difference, the difference between $\ce{O}$ and $\ce{H}$ is more than that between $\ce{O}$ and $\ce{C}$ so there should be even more negative charge at $\ce{O}$ in case of $\ce{H2O}$. In this way $\ce{H2O}$ should be less acidic than $\ce{CH3CH2-OH}$. This is also the reason that $+M$ (electron donating ability) of $\ce{-OH}$ (which is having more negative charge at $\ce{O}$) is greater than the $+M$ of $\ce{-OR}$.
\begin{array} {|l|r|r|}
\hline
  &amp;\mathrm{pK_a~in~water} &amp;\mathrm{pK_a~in~DMSO}\\
\hline
\ce{H2O} &amp;15.74 &amp;31.4\\
\hline
 \ce{CH3CH2OH} &amp;16 &amp;29.8 \\
\hline
\end{array}

As we can see from the $pK_a$ data water is slightly more acidic than ethanol in water but in DMSO ethanol is notably more acidic. In DMSO hydrogen bonding is minimal so the acidity is primarily dependent on the acid and not the solvent it is in. 

However, in water the hydroxide ion is hugely stabilised by hydrogen bonding (the ethoxide is as well but not nearly to the same extent) so this accounts for the greater acidity of water in this situation.

http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf
http://www.chem.wisc.edu/areas/reich/pkatable/index.htm