pH of a saturated solution of Ba(OH)_2 is 12. The value of solubility product ( K_sp ) of Ba(OH)_2 is: - Chemistry MCQ Chemistry | ExamTest.live

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pH of a saturated solution of Ba(OH)_{2} is 12. The value of solubility product (K_{s p}) of Ba(OH)_{2} is:

A $5.0 \times 1 0^{- 6}$
B $3.3 \times 1 0^{- 7}$
C $5.0 \times 1 0^{- 7}$
D $4.0 \times 1 0^{- 6}$

Solution & Step-by-step Explanation

1. Calculate [OH^{-}] from pH: pH = 12 ⟹ pOH = 14 - 12 = 2 [OH^{-}] = 1 0^{- pOH} = 1 0^{- 2} M 2. Relation between solubility (x) and ion concentration for Ba(OH)_{2} : Ba(OH)_{2} ⇌ Ba^{2 +} + 2 OH^{-} If solubility is x, then [Ba^{2 +}] = x and [OH^{-}] = 2 x . 2 x = 1 0^{- 2} ⟹ x = 5 \times 1 0^{- 3} M 3. Calculate K_{s p} : K_{s p} = [Ba^{2 +}] [OH^{-}]^{2} = x (2 x)^{2} = 4 x^{3} K_{s p} = (5 \times 1 0^{- 3}) \times (1 0^{- 2})^{2} K_{s p} = 5 \times 1 0^{- 3} \times 1 0^{- 4} = 5 \times 1 0^{- 7}

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pH of a saturated solution of Ba(OH)_{2} is 12. The value of solubility product (K_{s p}) of Ba(OH)_{2} is:

5.0 \times 1 0^{- 6}

3.3 \times 1 0^{- 7}

5.0 \times 1 0^{- 7}

4.0 \times 1 0^{- 6}

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pH of a saturated solution of $Ba(OH)_{2}$ is 12. The value of solubility product ( $K_{s p}$ ) of $Ba(OH)_{2}$ is:

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pH of a saturated solution of Ba(OH)2​ is 12. The value of solubility product (Ksp​) of Ba(OH)2​ is:

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pH of a saturated solution of $Ba(OH)_{2}$ is 12. The value of solubility product ( $K_{s p}$ ) of $Ba(OH)_{2}$ is: