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Lewis Theory

A 0.1 M aqueo...

Question

A 0.1 M aqueous solution of $K O H$ is slowly added to $100$ ml $0.1$ M aqueous solution of weak dibasic acid $H_{2} A .$ $(K_{a_{1}} = 10^{- 5}, K_{a_{2}} = 10^{- 9} f o r H_{2} A)$
Select the correct statement(s) among the following.

The

p H

after addition of

100

ml of

K O H

will be

7

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The

p H

after addition of

150

ml of

K O H

will be

9.

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If indicator

X, Y

Z

have

p H

range

(4 - 6),

(5.5 - 8)

(9 - 12)

respectively, then the most appropriate choice of indicators for Ist and

2^{n d}

equivalence point will be

X

Z

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Initial

p H

of the acid is

3

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Solution

The correct options are
B The $p H$ after addition of $150$ ml of $K O H$ will be $9.$
C Initial $p H$ of the acid is $3$ .
D The $p H$ after addition of $100$ ml of $K O H$ will be $7$ .
$H_{2} A + K O H \to H A^{-} + K^{+} + H_{2} O$
$100 \times 0.1 100 \times 0.1$
$= 10 m m o l$
$= 10 m m o l$
$H A^{-} + K O H \to A^{2 -} + K^{+} + H_{2} O$
$= 10 m m o l 50 \times 0.1$
$= 10 - 5$
$= 5 m m o l$
$= 5 m m o l$
(A) The solution now contains only $H A^{-}$ . So, the pH = $\frac{p K_{a 1} + p K_{a 2}}{2} = \frac{5 + 9}{2} = 7$ .
(B) After addition of $150$ ml $K O H$ , the second neutralisation step is $50$ % complete.
Hence solution contains 5 \:m mol of $H A^{-}$ and 5\: m mol of $A^{2 -}$ . So, it is a buffer.
$\Rightarrow p H = p K_{a 2} + l o g \frac{5}{5}$
$p H = 9$ .
(C) At the end point of first reaction, pH = 7. Hence, indicator Y is suitable.
(D) The pH of the initial acid is
$p H$ = $\frac{1}{2} (p K_{a 1} - l o g C)$
$= v \frac{1}{2} (5 - l o g 0.1) = 3$ .

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A 0.1 M aqueous solution of KOH is slowly added to 100 ml 0.1 M aqueous solution of weak dibasic acid H2A. (Ka1=10−5,Ka2=10−9forH2A)Select the correct statement(s) among the following.

A 0.1 M aqueous solution of $K O H$ is slowly added to $100$ ml $0.1$ M aqueous solution of weak dibasic acid $H_{2} A .$ $(K_{a_{1}} = 10^{- 5}, K_{a_{2}} = 10^{- 9} f o r H_{2} A)$
Select the correct statement(s) among the following.