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{p}{H} of saturated solution of {B}{a}{({O}{H})}_{{{2}}} is {1

Question

$p H$ of saturated solution of $B a (O H)_{2}$ is $12$ . The value of solubility product $(K_{s p})$ of $B a (O H)_{2}$ is

3.3 \times 10^{- 7}

5.0 \times 10^{- 7}

4.0 \times 10^{- 6}

5.0 \times 10^{- 6}

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p H

of saturated solution of

B a (O H)_{2}

12

. The value of solubility product

(K_{s p})

B a (O H)_{2}

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

Given, pH of

B a (O H)_{2} = 12

∴ p O H = 14 - p H

= 14 - 12 = 2
We know that,

p O H = - lo g [O H^{-}]

2 = - lo g [O H^{-}]

[O H^{-}]

= antilog (-2)

[O H^{-}] = 1 \times 10^{- 2}

B a (O H)_{2}

dissolves in water as

S mol L^{- 1} B a (O H)_{2} (S) \Leftrightarrow S B a^{2 +} + 2 S 2 O H^{-}

∴ [O H^{-}] = 2 S = 1 \times 10^{- 2}

S = \frac{[ O H ^{-} ]}{2} [B a^{2 +} = S]

[B a^{2 +}] = \frac{[ O H ^{-} ]}{2} = \frac{1 \times 10 ^{- 2}}{2}

K_{s p} = [B a^{2 +}] [O H^{-}]^{2}

= (\frac{1 \times 10 ^{- 2}}{2}) (1 \times 10^{- 2})^{2}

= 0.5 \times 10^{- 6} = 5 \times 10^{- 7}

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Question Text	$p H$ of saturated solution of $B a (O H)_{2}$ is $12$ . The value of solubility product $(K_{s p})$ of $B a (O H)_{2}$ is
Answer Type	Text solution:1
Upvotes	150