the molar solubility of Zn(OH)2 is 5.7x 10^-3 mol/L at a certain temperature. Calculate the value of Ksp for Zn(OH)2 at this temperataure

Ksp = 7.4x10⁻⁷

Explanation:

Molar solubility of a substance is defined as the amount of moles of that can be dissolved per liter of solution.

Ksp of Zn(OH)₂ is:

Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻

Ksp = [Zn²⁺] [OH⁻]²

And the molar solubility, X, is:

Zn(OH)₂(s) ⇄ Zn²⁺ + 2OH⁻

                 ⇄ X + 2X

Because X are moles of substance dissolved.

Ksp = [X] [2X]²

Ksp = 4X³

As molar solubility, X, is 5.7x10⁻³mol/L:

Ksp = 4X³

Ksp = 4 (5.7x10⁻³mol/L)³

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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