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 $\ce{-OR}$.
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