The ratio of `pH` of solution `(1)` containing 1 mole of` CH_(3)COONa` and 1 mole of `HCl` and solution `(II)` containing 1 mole of `CH_(3)COONa` and 1 mole of acetic acid in one litre is `:`

Explanation

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`.