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JEE Main levelSingle correct optionDifficulty 2 of 3ie012

What is the pH at 298 K of $0.02 M$ $Ba (OH) X_{2}$ (assume complete dissociation; $lo g 2 = 0.301$ )?

A $1.40$
B $12.30$
C $11.60$
D $12.60$

Answer: D

Ba (OH) X_{2}

gives two

OH X^{-}

per formula unit, so

[OH X^{-}] = 0.04 M

.

pOH = - lo g (4 \times 1 0^{- 2}) = 2 - 0.602 = 1.398

, hence

pH = 14 - 1.398 = 12.60

.

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