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Vikki [24]
2 years ago
9

The Ksp of mercury(II) hydroxide, Hg(OH)2, is 3.60×10^−26 . Calculate the solubility of this compound in grams per liter.

Chemistry
1 answer:
Stells [14]2 years ago
8 0

The solubility of the compound is obtained as 9.5 * 10^-14 M.

<h3>What is the solubility product?</h3>

The solubility product refers to the equilibrium constant that shows the extent to which a thing is soluble in water, Now we can write the solubility product as;

Ksp = [Hg^2+] [2OH^-]^2

Let  [Hg^2+]  =  [OH^-] = x

Then I can write;

Ksp = x * (2x)^2

Ksp = 4x^3

When x = solubility

x = ∛Ksp/4

Ksp =  3.60×10^−26

x =  ∛ 3.60×10^−26/4

x = 9.5 * 10^-14 M

Learn more about Ksp:brainly.com/question/27132799

#SPJ1

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