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Calculate the molar solubility of zinc hydroxide whose Ksp is 7. 7*10- 17 in a 0. 267M solution of iron(ii) hydroxide

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User LStarky
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1 Answer

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To solve this problem, we need to write the balanced equation for the dissolution of zinc hydroxide in water, and then write the expression for the solubility product constant, Ksp. The balanced equation is:

Zn(OH)2(s) ⇌ Zn2+(aq) + 2OH-(aq)

The Ksp expression is:

Ksp = [Zn2+][OH-]^2

We are also given that the concentration of iron(II) hydroxide, Fe(OH)2, is 0.267 M. We can use the concentration of hydroxide ions produced by the iron(II) hydroxide to find the concentration of hydroxide ions from the zinc hydroxide, since they will both be in equilibrium with the same hydroxide ions.

The balanced equation for the dissolution of iron(II) hydroxide is:

Fe(OH)2(s) ⇌ Fe2+(aq) + 2OH-(aq)

The concentration of hydroxide ions produced by the iron(II) hydroxide is twice the concentration of the iron(II) ions, or [OH-] = 2[Fe2+]. Substituting this into the Ksp expression for zinc hydroxide, we get:

Ksp = [Zn2+](2[Fe2+])^2

We can then solve for the molar solubility of zinc hydroxide, [Zn2+], by plugging in the values for Ksp and [Fe2+]:

7.7 x 10^-17 = [Zn2+](2 x 0.267 M)^2

Solving for [Zn2+], we get:

[Zn2+] = 7.7 x 10^-17 / (2 x 0.267 M)^2 = 5.12 x 10^-20 M

Therefore, the molar solubility of zinc hydroxide in a 0.267 M solution of iron(II) hydroxide is 5.12 x 10^-20 M.
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User Jibreel
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7.9k points
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