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InterviewSolution
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The pH of basic buffer mixtures is given by pH `= pK_(a) + log.(["Base"])/(["Salt"])`, whereas pH of acidic buffer mixtures is given by : pH `= pK_(1) + log.(["Salt"])/(["Acid"])`. Addition of little acid or base although shows no change in pH for all practical purposes , but since the ratio `(["Base"])/(["Salt"])` for `(["Salt"])/(["Acid"])` changes, a slight decrease or increase in pH results. The amount of `(NH_(4))_(2)SO_(4)` to be added to `500 mL` of 0.01 M `NH_(4)OH` solution (`pK_(a)` for `NH_(4)^(+)` is 9.26) to prepare a buffer of pH 8.26 isA. 0.05 moleB. 0.025 moleC. 0.10 moleD. 0.005 mole |
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Answer» Correct Answer - B `pH = pK_(a) + log.(["Base"])/(["Salt"]) rArr ["Base"] = (0.01 xx 500)/(500) = 0.01` `[NH_(4)^(+)] = (a xx 2)/(500)`, Let a millimole of `(NH_(4))_(2)SO_(4)` are added. `:. [Salt] = [NH_(4)^(+)]`. `pH = 9.26 + log [(0.01)/(2a//500)]` `8.26 = 9.26 + log.(0.01 xx 500)/(2a) :. a = 25` `:.` Mole of `(NH_(4))_(2)SO_(4)` added `= 0.025`. |
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