(a) When 100 ml of 0.1 M NaCN solution is titrated with 0.1 M HCI solution the variation of pH of solution with volume of HCI added will be :
(b) Variation of degree of dissociation $α$ with concentration for a weak electrolyte at a particular temperature is best represnted by :
(c) M acetic acid solution is titrated against M NaOH solution. the difference in pH between 1/4 and 3/4 stages of neutralization of acid will be 2 log 3.

T, F, T

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F, F, F

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T, T, T

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F, T, F

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Solution

The correct option is A $T, F, T$
(a) Initially pH will decrease fast, then slowly due to buffer formation and then will decrease fast as buffer action diminishes.

(b) for a weak electrolyte
$K_{a} = \frac{C α^{2}}{(1 - α)}$ when $α << 1 t h e n, α = \sqrt{\frac{k_{a}}{C}}$
as C increases $\Rightarrow α$ decreases
as C is tending to zero $\Rightarrow α$ will be unity

(c) At 1/4th neutralisation
$C H_{3} C O H + N a O H \to C H_{3} C O O N a + H_{2} O$
$(0.1 \times \frac{3}{4}) (0.1 \times \frac{1}{4})$
$p H = p K_{a} + l o g \frac{[C H_{3} C O O^{-}]}{[C H_{3} C O O H [} = p K_{a} + l o g (\frac{1}{3})$

At 3/4th neutralisation
$p H = p K_{a} + l o g 3$
So difference in pH = $Δ (p H) = l o g 3 - l o g \frac{1}{3} = 2 l o g 3$

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