SATHEE: Equilibrium - Result Question 76

Equilibrium - Result Question 76

####80. $p H$ of a 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 :

(a) $3.3 \times 10^{- 7}$

(b) $5.0 \times 10^{- 7}$

(d) $5.0 \times 10^{- 6}$

[2012]

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Solution:

(b) Given $p H = 12$

or $[H^{+}] = 10^{- 12}$

Since, $[H^{+}] [O H^{-}] = 10^{- 14}$

$∴ [O H^{-}] = \frac{10^{- 14}}{10^{- 12}} = 10^{- 2}$

$\begin{aligned} B a (O H)_{2} \to B a^{2 +} + 2 O H^{-} \\ [O H^{-}] = 10^{- 2} \\ 2 s = 10^{- 2} \\ s = \frac{10^{- 2}}{2} = 5 \times 10^{- 3} m, K_{s p} = s \cdot (2 s)^{2} \\ \Rightarrow K_{s p} = 5 \times 10^{- 7} \end{aligned}$

Equilibrium - Result Question 76

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Equilibrium - Result Question 76

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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