(a)When 100 ml of 0.1M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be : (b)Variation of degree of dissociation alpha with concentration for a weak electrolyte at a particular temperature is best represented by : ( c)0.1 M acetic acid solution is titrated against 0.1 M NaOH solution. The difference in pH between 1/4 and 3/4 stages of neutralization of acid will be 2 log 3.

(a)When 100 ml of 0.1M NaCN solution is titrated with 0.1 M HCl soluti

1.	(a)When 100 ml of 0.1M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be : (b)Variation of degree of dissociation alpha with concentration for a weak electrolyte at a particular temperature is best represented by : ( c)0.1 M acetic acid solution is titrated against 0.1 M NaOH solution. The difference in pH between 1/4 and 3/4 stages of neutralization of acid will be 2 log 3.
Answer» T,F,T F,F,F T,T,T F,T,F Solution :(a)Initially pH will decrease fast, then slowly DUE to buffer FORMATION and then will decrease fast as buffer action diminishes. (b) For a weak electrolyte `K_a=(Calpha^2)/((1-alpha))` when `alphaltlt 1` then `alpha=sqrt(k_a/C)` as C increases `implies` `alpha` decreases as C is tending to ZERO `implies` `alpha` will be unity ( c) At `1//4^(th)` neutralisation `CH_3COOH+NaOHtoCH_3COONa+H_2O` `(0.1xx3/4) " " (0.1xx1/4)` `pH=pK_a+"LOG" ([CH_3COO^(-)])/([CH_3COOH])=pK_a+ "log" (1/3)` At `3//4^(th)` neutralisation `pH=pK_a+"log" 3` so difference in pH=`DELTA(pH)="log"3-"log"1/3=2"log"3`