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

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 .

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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