Question

The solubility product of $Mg(OH)_{2}$ is $1.2\times10^{-11}$ at $25^{0}C$. Calculate solubility of $Mg(OH)_{2}$ in water in grams/Lit. (Mol. wt. of $Mg(OH)_{2}$ is 58).

Answer 1

Answer:

Content:

Let solubility be S g/mol

Mg(OH)2------->Mg2+ + 2OH-

Solubility of Mg2+ is S

Solubility of OH- is 2S

Ksp=S*(2S)^2=4S^3

So,

1.2*10^-11=4S^3

S=1.44*10^-4 g/mol

We know 1 mole contains 22.4l by volume.

So, solubility of magnesium hydroxide in g/l is (1.44*10^-4)*22.4=3.225*10^-3 g/l

Answer 2

Answer:

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