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Solubility product constant

At 25^{} {C}, the solubility product of {Mg}({OH})_{2} is 1.0

Question

Medium

Solving time: 3 mins

At $2 5^{\circ} C$ , the solubility product of $Mg (OH)_{2}$ is $1.0 \times 1 0^{- 11}$ . At which $pH$ , will $Mg^{2 +}$ ions start precipitating in the form of $Mg (OH)_{2}$ from a solution of $0.001 M Mg^{2 +}$ ions?

Views: 6,118 students

Updated on: Dec 26, 2022

Book:

Arihant Chemistry JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2010

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Liam Discussed

2 5^{\circ} C

, the solubility product of

Mg (OH)_{2}

1.0 \times 1 0^{- 11}

. At which

pH

, will

Mg^{2 +}

ions start precipitating in the form of

Mg (OH)_{2}

from a solution of

0.001 M Mg^{2 +}

ions?

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

Mg (OH)_{2} ⇌ Mg^{2 +} + 2 OH^{-}

K_{sp} = [Mg^{2 +}] [OH^{-}]^{2}

[OH^{-}] = \frac{K _{sp}}{[ Mg ^{2 +}}] = \frac{1 \times 1 0 ^{- 11}}{0.001}

pOH = - lo g [OH^{- 1}] = - lo g [1 0^{- 4}]

pOH = 4 and pH = 10

∴ pH = 14 - pOH = 14 - 4 = 10

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Question Text	At $2 5^{\circ} C$ , the solubility product of $Mg (OH)_{2}$ is $1.0 \times 1 0^{- 11}$ . At which $pH$ , will $Mg^{2 +}$ ions start precipitating in the form of $Mg (OH)_{2}$ from a solution of $0.001 M Mg^{2 +}$ ions?
Updated On	Dec 26, 2022
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 4
Upvotes	529
Avg. Video Duration	6 min

At $2 5^{\circ} C$ , the solubility product of $Mg (OH)_{2}$ is $1.0 \times 1 0^{- 11}$ . At which $pH$ , will $Mg^{2 +}$ ions start precipitating in the form of $Mg (OH)_{2}$ from a solution of $0.001 M Mg^{2 +}$ ions?

Text solutionVerified

Filo tutor solutions (4)

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At 25∘C, the solubility product of Mg(OH)2​ is 1.0×10−11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2​ from a solution of 0.001MMg2+ ions?

Text solutionVerified

Filo tutor solutions (4)

Questions from JEE Mains 2010 - PYQs

Practice questions from Arihant Chemistry JEE Main Chapterwise Solutions (2019-2002) (Arihant)

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At $2 5^{\circ} C$ , the solubility product of $Mg (OH)_{2}$ is $1.0 \times 1 0^{- 11}$ . At which $pH$ , will $Mg^{2 +}$ ions start precipitating in the form of $Mg (OH)_{2}$ from a solution of $0.001 M Mg^{2 +}$ ions?