When $2\mathrm{.}0$ mole of acetic acid is diluted with water to a volume of $1$L$,$ a small percentage of the acid ionizes to form acetate ion and hydroniun ion. CH$3$CO$2$H $($aq$)+$ H$2$O $($l$)=$ CH$3$CO$2-($aq$)+$ H$3$O$+($ aq$).$ At equilibrium, $0\mathrm{.}30\%$ of acetic acid ionizes at $25$ degrees C$.$ What percentage of the acid ionizes when $0\mathrm{.}50$ mole of acetic acid is diluted with water to $1$L at $25$ degrees C$?$

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To determine the percentage of acetic acid that ionize... View the full answer