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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_(a)+"log"(["Salt"])/(["Acid"])`. Addition of little acid or base although shows no appreciable change in Ph for all practical purposes, but sicne the ratio `(["Base"])/(["Salt"])` or `(["Salt"])/(["Acid"])` changes, a slight decrease or increase in pH results. The ratio of pH of solution (I) containing 1 mole to pH of solution (II) containing 1 mole of `CH_(3)COONa` and 1 mole of acetic in one litre is :A. `1:2`B. `2:1`C. `1:3`D. `3:1` |
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Answer» `{:(CH_(3)COONa,+,HCl,rarr,CH_(3)COOH,+,NaCl),(1,,1,,0,,0),(0,,0,,1,,1):}` `:.[CH_(3)COOH]=(1)/(1)=1` `:.[H^(+)]=Calpha=Csqrt((K_(a))/(C))=sqrt(K_(a).C)=sqrt(K_(a))` or `pH_(1)=-(1)/(2)logK_(a)=(1)/(2)pK_(a)` `underset(1)(CH_(3)COOH)+underset(1)(CH_(3)COONa)` `:.pH=pK_(a)+(log)(1)/(1)` `pH_(2)=pK_(a)` `(pH_(1))/(pH_(2))=(1)/(2)` |
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