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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 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`) isA. 50 mLB. 25 mLC. 20 mLD. 10 mL |
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Answer» Correct Answer - B 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`.` |
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