SATHEE: Equilibrium - Result Question 81

Equilibrium - Result Question 81

####85. If $p H$ of a saturated solution of $B a (O H)_{2}$ is 12 , the value of its $K_{(s p)}$ is :

[2010]

(a) $4.00 \times 10^{- 6} M^{3}$

(b) $4.00 \times 10^{- 7} M^{3}$

(d) $5.00 \times 10^{- 7} M^{3}$

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Solution:

(d) $B a (O H)_{2} (s) ⟶ B a^{2 +} (a q) + 2 O H^{-}$ (aq) $p H = 12$ or $p O H = 2$

$[O H^{-}] = 10^{- 2} M$

$B a (O H)_{2} ⟶ B a^{2 +} + 2 O H^{-}$

$0.5 \times 10^{- 2} 10^{- 2}$

$[∴ .$ Concentration of $B a^{2 +}$ is half of $. O H^{-}]$

$\begin{aligned} K_{s p} & = [B a^{2 +}] [O H^{-}]^{2} \\ = [0.5 \times 10^{- 2}] [1 \times 10^{- 2}]^{2} \\ = 0.5 \times 10^{- 6} = 5 \times 10^{- 7} M^{3} \end{aligned}$

Equilibrium - Result Question 81

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Equilibrium - Result Question 81

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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