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 volume of 0.2 M NaOH needed to prepare a buffer of pH 4.74 with 50 ml of 0.2 M acetic acid (pK_(b) of CH_(3)COO^(-) = 9.26) 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 volume of 0.2 M NaOH needed to prepare a buffer of pH 4.74 with 50 ml of 0.2 M acetic acid (pK_(b) of CH_(3)COO^(-) = 9.26) is
Answer» 50 mL 25 mL 20 mL 10 mL Solution :Let V mL of NaOH be needed to give `CH_(3)COONa`. `{:(NaOH+,CH_(3)COO,rarr,CH_(3)COONa+,H_(2)O),(0.2 xx V,50 xx 0.2,,""0," "0),(-,[10-0.2V],,""0.2V,0.2 V):}` `:. pH = pK_(a) + LOG.(["Salt"])/(["ACID"]) = pK_(W) - pK_(b) + log.(["Salt"])/(["Acid"])` `= 14 - 9.26 + g.(["Salt"])/(["Acid"])` `= 14-9.26 + log.([(0.2V)/(50+V)])/([(10-0.2 V)/(50+V)])` `4.74 = 4.74 + log [(0.2V)/(10 - 0.2 V)] :. V = (10)/(0.4) = 25 mL`.`