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Calculate the concentration of `H_(3)O^(+)` ions in a mixture of 0.02 M acetic acid and 0.2 M sodium acetate. Given that the ionization constant `(K_(a))` for acetic acid is `1.8xx10^(-5)`. |
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Answer» `{:(,CH_(3)CO OH +H_(2)O,hArr,CH_(3)CO O^(-) +, H_(3)O^(+),),("Initial moles",0.02,,0,0,),("Moles at eqm.",0.02-x,,x,x,),(,CH_(3)CO ONa,rarr,0,0,),("Initial moles",0.2,,0,0,),("Moles at eqm.",0,,0.2,0.2,),(,,,,,):}` Thus, in the mixture solution, `[CH_(3)CO O^(-)]=0.2+x~=0.2 M (CH_(3)CO O^(-)" are obtained mainly from" CH_(3)CO ONa," therefore," x lt lt 0.2)` `[CH_(3)CO OH ] = 0.02 - x ~= 0.02 M` `K_(a)=([CH_(3)CO O^(-)][H_(3)O^(+)])/([CH_(3)CO OH]), i.e., 1.8xx10^(-5)=(0.2xx[H_(3)O^(+)])/(0.02)` or`[H_(3)O^(+)]=(1.8xx10^(-5)xx0.02)/(0.2) = 1.8xx10^(-6)M` Note. The given mixture is a buffer as discussed later in Art. 7.32 . |
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