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MolSynapse / Question bank / Equivalent Concept and Titration / Acid-base titrations and double titration (NaOH, Na2CO3, NaHCO3)

JEE Advanced levelSingle correct optionDifficulty 3 of 3eq053

$25.0 mL$ of a saturated filtered solution of $Ca (OH) X_{2}$ requires $20.0 mL$ of $0.0400 M$ $HCl$ for neutralisation. Assuming complete dissociation, the solubility product of $Ca (OH) X_{2}$ from this data is:

A $4.10 \times 1 0^{- 6}$
B $1.64 \times 1 0^{- 5}$
C $3.28 \times 1 0^{- 5}$
D $5.12 \times 1 0^{- 5}$

Answer: B

OH X^{-}

in

25.0 mL = 0.0400 \times 20.0 = 0.800

mmol, so

[OH X^{-}] = 0.032 M

and

[Ca X^{2 +}] = 0.016 M

.

K_{s p} = [Ca X^{2 +}] [OH X^{-}]^{2} = 0.016 \times (0.032)^{2} = 1.64 \times 1 0^{- 5}

.

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