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1.0 M hydroxide will form a precipitate with copper(II) cation but 6.0 M hydroxide will dissolve the solid. Explain using equations and Ksp and Kf arguments. (Hint: Consider that you have two reactions happening: the Ksp reaction of the solid, and the Kf reaction to form the complex ion. What happens when those two reactions combine

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Answer:

Explanation:

Low concentration of hydroxide ion forms a precipitate

Cu²⁺+ 2OH⁻ --------->Cu(OH)₂〈s〉

Ksp= [Cu²⁺][OH⁻]² =4.8* 10⁻²⁴ at 25⁰C

At higher concentration 6M hydroxide of complex ion is formed where Cu+ is ligand

Cu(OH)₂〈s〉 + 2OH⁻ ----------->[Cu(OH₄)]²⁻

Kf = 1.3 * 10¹⁶

Kp>> Ksp

Therefore, at lower concentration, a precipitate will form .

When both these reactions combine at low concentration of hydroxide ion forms a precipitate.

Cu²⁺+ 2OH⁻ --------->Cu(OH)₂〈s〉

Ksp = [Cu²⁺][OH⁻]²

Ksp =4.8* 10⁻²⁴ at 25⁰C

At higher concentration 6.0 M hydroxide of complex ion is formed where Cu+ is ligand

Solubility product:

The solubility product is a kind of equilibrium constant and its value depends on temperature. Ksp usually increases with an increase in temperature due to increased solubility.

Cu(OH)₂〈s〉 + 2OH⁻ ----------->[Cu(OH₄)]²⁻

Kf = 1.3 * 10¹⁶

Kp>> Ksp

Therefore, at lower concentration, a precipitate will form.

Find more information about Solubility product here:

brainly.com/question/9807304

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