SATHEE: Equilibrium Question 904

Equilibrium Question 904

Question: The solubility product of $M g {(O H)}_{2}$ is $10^{- 14}$ . The solubility of $M g {(O H)}_{2}$ in a buffer solution of $p H = 8$ is

Options:

A) $10^{- 8}$

B) $10^{- 6}$

C) $10^{- 2}$

D) $10^{- 4}$

Show Answer

Answer:

Correct Answer: C

Solution:

$M g {(O H)}_{2}, M g^{2 +}, + 2 O H^{-}$

$K_{s p} = [M g^{2 +}] {[O H^{-}]}^{2}$

$1 \times 10^{- 14} = [M g^{2 +}], {[10^{- 6}]}^{2}$

$∵$ $p H = 8$

$∴$ $[H^{+}] = 10^{- 8}, m o l, L^{- 1}$ . and $[O H^{-}] = 10^{- 6} m o l L^{- 1}]$

$[M g^{2 +}] = \frac{10^{- 14}}{10^{- 12}} = 10^{- 2} m o l, L^{- 1}$

Solubility $[M g^{2 +}] = 10^{- 2}, m o l L^{- 1}$

Equilibrium Question 904

Question: The solubility product of $M g {(O H)}_{2}$ is $10^{- 14}$ . The solubility of $M g {(O H)}_{2}$ in a buffer solution of $p H = 8$ is

Options:

Answer:

Solution:

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Equilibrium Question 904

Question: The solubility product of Mg(OH)2 is 10−14 . The solubility of Mg(OH)2 in a buffer solution of pH=8 is

Options:

Answer:

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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Question: The solubility product of $M g {(O H)}_{2}$ is $10^{- 14}$ . The solubility of $M g {(O H)}_{2}$ in a buffer solution of $p H = 8$ is