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The `pH` of a solution of weak base at neutralisation with strong acid is `8. K_(b)` for the base isA. `1.0 xx 10^(-4)`B. `1.0 xx 10^(-6)`C. `1.0 xx 10^(-8)`D. None of these |
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Answer» Correct Answer - B At half neutralisation, `[B^(o+)] = [BOH] , ["Salt"] = ["base"]` `pOH = pK_(b) + "log" ([B^(o+)])/([BOH])` `pOH = pK_(b), (pH = 8, pOH = 6)` `:. pK_(b) = 6, K_(b) = 1 xx 10^(-6)` |
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