Which of the following shows the correct order of decreasing acidity?
(Here $P h -$ indicates phenyl group)

P h C O_{2} H > P h S O_{3} H > P h C H_{2} O H > P h O H

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P h S O_{3} H > P h O H > P h C O_{2} H > P h C H_{2} O H

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P h C O_{2} H > P h O H > P h C H_{2} O H > P h S O_{3} H

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P h S O_{3} H > P h C O_{2} H > P h O H > P h C H_{2} O H

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Solution

The correct option is D $P h S O_{3} H > P h C O_{2} H > P h O H > P h C H_{2} O H$
$P h C H_{2} O H$ is least acidic than others because its conjugate base is not stabilised by any resonating structures like others.
Conjugate bases of $P h S O_{3} H$ and $P h C O O H$ have equivalent structures but conjugate base of $P h O H$ has non-equivalent resonating structures.
Also, the negative charge is delocalised on less electronegative carbon atom while in conjugate bases of $P h S O_{3} H$ and $P h C O O H$ , the negative charge is delocalised on highly electronegative oxygen atoms which makes $P h S O_{3} H$ and $P h C O O H$ stronger acid than $P h O H$ , due to increased stability of their conjugate base.
Now,
Conjugate $P h S O_{3} H$ has 3 equivalent resonating structures while $P h C O O H$ has one less i.e. 2 equivalent resonating structures as shown.

Higher the number of equivalent structure more will be the stability of conjugate base. Hence $P h S O_{3} H$ has most stable conjugate base.
Stability of conjugate base $\propto$ Strength of acid
Therefore, the decreasing order of acidity is
$P h S O_{3} H > P h C O_{2} H > P h O H > P h C H_{2} O H$

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