Question

What is the molar solubility of Mg(OH)2 in a basic solution with a pH of 12.00? Ksp for Mg(OH)2 is 5.6 × 10-12.

What is the molar solubility of Mg(OH)2 in a basic solution with a pH of 12.00? Ksp for
Mg(OH)2 is 5.6 × 10-12.
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Answer #1
Concepts and reason

Molar solubility is the number of moles of compound or substance that can dissolved per liter of solution before solution get saturated and it depends on the stoichiometry of the dissolution reaction and it can be calculated by the solubility product constant. pH is a concentration of hydrogen ion in a solution. And it indicate solution are acidic basic or neutral means where solution with high concentration of hydrogen ions will be less pH value with higher acidic and with low concentration of hydrogen ions have high pH value means higher basic solutions.

Fundamentals

Solubility product constant and molar solubility calculated by the concentration of reaction let a compound reaction;

X,Y()2X(aq) +Y(aq)

Ksp =[X][Y]
where, Y = xso, X = 2x

Now, we can find x that is molar solubility of that compound.

The ionization of Mg(OH)2
; and K. = 5.6x10-12

Mg(OH),(s)
Mg²+ (aq) +20H (aq)

pH = 12 so,

pOH = 14-12 = 2

pOH = 14 -12 = 2
[OH-]=10 2 = 0.01M

So,

Mg(OH)2

initial

-

0

0.01 M

Change

-

x

2x+0.01 M

At equilibrium

-

x

2x+0.01 M

Solubility product expression:

[OH-]=102 = 0.01M
K. = [Mg2+ ][20H-1
K =[x][2x+0,01M ]
2x <<0.01M so, 2x can be neglected
K =[x][0.0001]
5.6x10-12 /0.0001 =

Ans:

Molar solubility of Mg(OH)2
= 5.6x10 ⓇM

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