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When a salt reacts with water to form a...

When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations :
`pH = 1/2 [pK_(w) +pK_(a) + logc] ` (for salt of weak acid and strong base .)
`pH = 1/2 [pK_(w) - pK_(b) - logc] ` (for salt of weak base and strong acid ) .
`pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] ` (for weak acid and weak base ).
where 'c' represents the concentration of salt .
When a weak acid or a weak base not completely neutralized by strong base or strong acid
respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation :
`pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"])`
Answer the following questions using the following data :
`pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14`
When 100 mL of `0.1` M `NH_(4)OH` is added to 50 mL of `0.1M` HCl solution , the pH is

Text Solution

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The correct Answer is:
`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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When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations : pH = 1/2 [pK_(w) +pK_(a) + logc] (for salt of weak acid and strong base .) pH = 1/2 [pK_(w) - pK_(b) - logc] (for salt of weak base and strong acid ) . pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] (for weak acid and weak base ). where 'c' represents the concentration of salt . When a weak acid or a weak base not completely neutralized by strong base or strong acid respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation : pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"]) Answer the following questions using the following data : pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14 0.001 M NH_(4)Cl aqueous solution has pH :

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