Q. If first dissociation of $X(OH)_{3}$ is 100% where as second dissociation is 50% and third dissociation is negligible then the $pH$ of $4 \times 10^{- 3}$ M $X(OH)_{3}$ is

A) 7.5
B) 9.54
C) 11.78
D) 13.25
Explanation

(c)
First dissociation
$X(OH)_{3} \rightarrow X(OH)_{2}^{+} + OH^{-}$
Second dissociation
$X(OH)_{2}^{+} \rightarrow X(OH)^{2 +} + OH^{-}$
Total $\left\lbrack OH^{-} \right\rbrack = 4 \times 10^{- 3} + 2 \times 10^{- 3}$
$= 6 \times 10^{- 3}$
$pOH = 3 - \log 6$
$= 3 - 0.78 = 2.22$
$pH = 14 - 2.22 = 11.78$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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