1.

Calcualte the `pH` at which an indicator with `pK_(b) = 4` changes colour.

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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