InterviewSolution
Saved Bookmarks
| 1. |
`K_(a)`and `K_(b)` are the dissociation constants of weak acid and weak base and `K_(w)` is the ions product of water. Match the pH stated in Column II with the solutions listed in Column I at `25^(@)C`. |
|
Answer» Correct Answer - `A rarr q; B rarr C rarr p; D rarr p, s` (A) `KCN + H_(2)O rarr KOH + HCN` (weak acid) `pH = (1)/(2) (pK_(w) + pK_(a) + log[CN^(-)])` `= (1)/(2)[14 + pK_(a) + log 0.1] = 6.5 + (1)/(2)pK_(a)` (B) `C_(6)H_(5)NH_(3)Cl + H_(2)O rarr C_(6)H_(5)NH_(3)OH + HCl` (Cataionic hydrolysis) `pH = (1)/(2)(pK_(a) - pK_(b) - log[C_(6)H_(5)NH_(3)^(+)])` `= (1)/(2)[14 + pK_(b) - log 0.1] = 7.5 + (1)/(2)pK_(b)` (C) KCl - Salt of strong acid (HCl) and strong base (KOH), hence no salt hydrolysis, pH = 7 (D) `CH_(3)COO^(-) + NH_(4)^(+) + H_(2)O rarr CH_(3)COOH + NH_(4)OH` `pH = (1)/(2)(pK_(w)+pK_(b) -pK_(b)) = (1)/(2)[14+pK_(a) - pK_(b)] = 7` (`pK_(a) = pK_(b)` in this case). |
|