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                                    0.2 M NaOH is titrated with 100 ml 0.2 M CH_(3) COOH in a conducitivity cell. The data obtained were plotted as following The pH of solution in conductivity cell is 9 at point B. Calculate pH of conductivity cell when 100 ml 0.1 M HCl added in resulting solution present in conducitivity cell at point B. | 
                            
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Answer»  `{:(,(0.2xx100)/(1000),(0.2xx100)/(1000)," "0," "0),(," "0,0,0.02," "0.02):}` `pH=9=(1)/(2)(pKw+pKa+log .(0.02)/(0.2))` ` 18=14+pKa-1` `pKa=5` `CH_(3)COONa(aq)+HCl(aq)rarrCH_(3)COOH(aq)+NaCl(aq)` `{:(,0.02,(100xx0.1)/(1000)=0.01,0,0),(,0.01," "0,0.01,0.01):}` Acidic buffer `pH=pKa+log.(CB)/(CA)` `pH=5+log.(0.01)/(0.01)` `ph=5`  | 
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