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Find the concentration of `H^(o+), HCO_(3)^(Θ)`, and `CO_(3)^(-2)` in a `0.01M` solution of carbonic acid if the `pH` of solution is `4.18`. `K_(1) = 4.45 xx 10^(-7),K_(2) = 4.69 xx 10^(-11)` |
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Answer» Given, `pH = 4.18 =- log [H^(o+)]` `:. [H^(o+)] = 6.61 xx 10^(-5) molL^(-1)` `H_(2)CO_(3) hArr H^(oplus)+HCO_(3)^(Theta)` `K_(1) = ([H^(o+)][HCO_(3)^( Θ)])/([H_(2)CO_(3)])` or `4.45 xx 10^(-7) = ([6.61xx10^(-5)][HCO_(3)^(Θ)])/([0.01])` or `[HCO_(3)^(Θ)] = 6.73 xx 10^(-5) molL^(-1)` Again for dissociation of `HCO_(3)^(Θ)`, we have `[HCO_(3)^(Theta)]hArr H^(oplus)+CO_(3)^(2-)` `K_(2)=([H^(o+)][CO_(3)^(2-)])/([HCO_(3)^(Θ)])` or `4.69 xx 10^(-11) = ([6.61 xx 10^(-5)][CO_(3)^(2)])/([6.73xx10^(-5)])` `[CO_(3)^(2-)] = 4.78 xx 10^(-11) molL^(-1)` |
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