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Characteristics of Equilibrium Constant

50 ml 0.1M Na...

Question

$50 m l 0.1 M N a O H$ is added to $50 m l$ of $0.1 M$ $C H_{3} C O O H$ solution, The $p H$ will be: (pKa ${C H}_{3} C O O H = 4.7447$ , Given $log 20 = 1.3$ )

4.7447

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9.2553

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8.7218

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n o n e o f t h e a b o v e

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Solution

The correct option is D $n o n e o f t h e a b o v e$ .
$N a O H + C H_{3} C O O H \to C H_{3} C O O N a + H_{2} O$
no. of mili equivalents = molarity $\times$ volume $\times$ n-factor
Meq. of $N a O H = 0.1 \times 60 \times 1 = 6$
Meq. of $C H_{3} C O O H = 0.1 \times 50 \times 1 = 5$
$N a O H + C H_{3} C O O H \to C H_{3} C O O N a + H_{2} O$
initial $6$ $5$
final $1$ $0$ $5$ $5$
In solution, we have strong base. so salt $C H_{3} C O O N a$ is neglected.
(SO) $B O H = - log [O H^{-}]$
no. of mili equivalent of $N a O H = 1$
so, $1 = n o r m a l i t y \times v o l u m e$
$1 = N \times 110$
$1 = N$
_____
$110$
$p O H = - log [1 / 110] = 2.04$
So, $p H = 11.96$

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