Calculate the ratio of pH of a solution containing 1 mole of CH_(3)CO Ona + 1mole of HCl per litre to that of a solution containing 1 mole of CH_(3)CO Ona + 1mole of CH_(3)CO OHper litre.

Calculate the ratio of pH of a solution containing 1 mole of CH_(3)CO

1.	Calculate the ratio of pH of a solution containing 1 mole of CH_(3)CO Ona + 1mole of HCl per litre to that of a solution containing 1 mole of CH_(3)CO Ona + 1mole of CH_(3)CO OHper litre.
Answer» SOLUTION :Case I. Calculation of pH of solution sontaining 1 mole of `CH_(3)CO ONA +1` mole of HCl per litre `{:(,CH_(3)CO ON a ,+,HCl,rarr,CH_(3)CO OH,+,NaCl ),("Initial moles",1 "mole",,1 "mole",,0,,0),("Moles after reaction ",0,,0,,1,,1):}` i.e., `[CH_(3)CO OH ] = 1 "mol"L^(-1)` `{:(,CH_(3)CO OH ,hArr,CH_(3)CO O^(-),+,H^(+),,),("Initial conc.",C "mol" L^(-1),,,,,,),("After dissociation",C - C alpha,,C alpha,,C alpha ,,):}` `:. [H^(+)]=C alpha . "But" alpha = sqrt((K_(a))/(C))` `:. [H^(+)]=C sqrt((K_(a))/(C))=sqrt(K_(a)C)=sqrt(K_(a))=K_(a)^(1//2)`(`:' C = 1 "mol" L^(-1)`) `:. - log [H^(+)]= - (1)/(2) log K_(a), i.e., (pH)_(1) = - (1)/(2)log K_(a)"" ...(i)` Case II. Calculation of pH of solution containing 1 mole of `CH_(3)CO ONa + 1 ` mole of `CH_(3)CO OH` per litre Applying Henderson equation, `(pH)_(2)=pK_(a) + log .(["Salt"])/(["ACID"]) = pK_(a)= - log K_(a)`...(ii) [[Salt]=[Acid]=1 mol `L^(-1)`] From equations (i) and (ii), `(pH)_(1)//(pH)_(2)=1//2`.