An organic monoprotic acid $[0.1 M]$ is titrated against $0.1 M$ $N a O H$ . By how much does the $p H$ change between one fourth and three fourth stages of neutralization? If at one-third stage of neutralization, the $p H$ is $4.45$ , what is the dissociation constant of the acid? Between what stages of neutralization may the $p H$ change by $2$ units?

0.6423,

p k_{a} = 2.43

\frac{10}{11} t h

and

\frac{1}{11} t h

stages of neutralization

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0.7693,

p k_{a} = 3.89

\frac{1}{11} t h

and

\frac{10}{11} t h

stages of neutralization

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0.9542,

p k_{a} = 4.751

\frac{1}{11} t h

and

\frac{10}{11} t h

stages of neutralization

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0.9852,

p k_{a} = 4.9

\frac{10}{11} t h

and

\frac{1}{11} t h

stages of neutralization

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Solution

The correct option is C $0.9542,$ $p k_{a} = 4.751$ , $\frac{1}{11} t h$ and $\frac{10}{11} t h$ stages of neutralization
(a) The $p H$ at one fourth neutralization,

${(p H)}_{1} = p k_{a} + log \frac{x / 4}{3 x / 4} = p k_{a} + log \frac{1}{3}$

The $p H$ at three fourth neutralization,

${(p H)}_{2} = p k_{a} + log \frac{3 x / 4}{x / 4} = p k_{a} + log 3$

$Δ$ $p H = {(p H)}_{2} - {(p H)}_{1} = 2 log 3$

$= 0.9542$

(b) $4.45 = p k_{a} + log \frac{x / 3}{2 x / 3} = p k_{a} - log 2$

$p k_{a} = 4.751$

(c) $Δ p H = 2$

${(p H)}_{1} = p k_{a} + 1$ , ${(p H)}_{2} = p k_{a} - 1$

For $p k_{a} + 1$ $\Rightarrow$ $\frac{[S]}{[A]} = 10$

$\frac{x}{a - x} = 10$ $\Rightarrow$ $x = 10 a - 10 x$

$x = \frac{10 a}{11}$

i.e., $\frac{10^{t h}}{11}$ stage

For $p k_{a} - 1$ $\Rightarrow$ $\frac{[S]}{[A]} = \frac{1}{10}$

$\frac{x}{a - x} = \frac{1}{10}$

$x = \frac{a}{11}$ i.e., $\frac{1}{11} t h$ stage

Hence, the correct option is $C$

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