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).


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