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 is

The pH of basic buffer mixtures is given by pH = pK_(a) + log.(["Base"

1.	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 is
Answer» 0.05 mole 0.025 mole 0.10 mole 0.005 mole Solution :`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`.