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Calculate the `H^(+)` ion concentration in 0.10 M acetic acid solution. Given that the dissociation constant of acetic acid in water is `1.8xx10^(-5)`.

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`{:(,CH_(3)CO OH+aq,hArr,CH_(3)CO O^(-) (aq),+,H^(+)(aq)),("Initial conc.",0.10 M ,,0M,,0M),("Concs. at eqm.",(0.10-x)M,,x M,,x M):}`
`K_(a)=([CH_(3)CO O^(-)(aq)][H^(+)(aq)])/([CH_(3)CO OH])`
`1.8xx10^(-5)=(x xx x)/(0.10-x) ~~(x^(2))/(0.10) or x^(2)=(1.8xx10^(-5))xx0.10=1.8xx10^(-6)`
or `x=sqrt(1.8xx10^(-6))=1.34xx10^(-3) M, i.e., [H^(+)]=1.34xx106(-3)M`

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