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Calcualte the `pH` at which an indicator with `pK_(b) = 4` changes colour. |
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Answer» `"Ind"^(Theta) +H_(2)O hArr "Hin" + overset(Theta)(O)H` At the colour change, `["Ind"^(Θ)] = ["Hin"]` `K_(b) = ([overset(Θ)OH]["Hin"])/(["Ind"^(Θ)]) = 1.0 xx 10^(-4) = [overset(Θ)OH]` and `pOH = 4.0` `:. pH = 10.00` |
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