Question

Find [Cu²⁺] in a solution saturated with Cu₄(OH)₆(SO₄) if [OH^-] is fixed at 6.5 x 10^-6 M. Note that Cu₄(OH)₆(SO₄) gives 1 mol of SO₄^2- for 4 mol Cu²⁺. Cu₄(OH)₆(SO₄)(s) <--> 4Cu²⁺ + 6OH^- + SO₄^2- Ksp = 2.3 x 10^-69

Answer 1

Let the concentration of Cu²⁺ be x M. we know that 1 mol of SO₄^2- give 4 mol of  Cu²⁺ . So concentration of SO₄^2- = x/4

We know, for a rxn: aA   bB + cC

K_sp = ( [B]^b [C]^c ) / [A]^a

Solids are not included when calculating equilibrium constant expressions, because their concentrations do not change the expression; any change in their concentrations are insignificant, and therefore omitted.

K_sp = [x⁴ X (x/4) X (6.5X10^-6) ⁶] / 1 = 2.3 X 10^-69

x⁵ = (2.3 X 10^-69 ) / (18854.7226 X 10^-36 ) = 1219.85 X 10^-40

x = 4.1425 X 10^-8 M = [Cu²⁺]

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