Find [Cu^2+] in a solution saturated with Cu_4(OM)_6(SO_4) If [OH^-] is fixed at

Question

Find [Cu^2+] in a solution saturated with Cu_4(OM)_6(SO_4) If [OH^-] is fixed at 1.8 times 10^-6 M. Note that Cu_4(OH)_6(SO_4) gives 1 mol of SO^2-_4 for 4 mol of Cu^2+. Cu_4(OH)_6(SO_4)(s) 4 Cu^2+ 6 OH^- + SO^2-_4 K_sp = 2.3 times 10^69 (a) From the solubility product of line ferrocyanide, Zn_2Fe(CN)_6, calculate the concentration of Fe(CN)^4-_6 in 0.13 mM ZnSO_4 saturated with Zn_2Fe(CN)_6. Assume that Zn_2Fe(CN)_6 is a negligible source of Zn^2+. (The K_sp Zn_2Fe(CN)_6 is 2.1 times 10^-16.) (b) What concentration of K_4Fe(CN)_6 should be in a suspension of solid Zn_2Fe(CN)_6 in water to give [Zn^2+] = 4.9 times 10^-7 M?

Explanation / Answer

For the given Ksp equations,

1) Ksp for (Cu)4(OH)6(SO4) = 2.3 x 10^-69

Ksp = [Cu2+]^4.[OH-]^6.[SO4^2-]

with,

[OH-] = 1.8 x 10^-6 M

we would have,

2.3 x 10^-69 = [Cu2+]^4.(1.8 x 10^-6)^6.(1.8 x 10^-6)

[Cu2+] = 1.05 x 10^-10 M as the concentration of Cu2+

2)

(a) Zn2Fe(CN)6 Ksp = 2.1 x 10^-16

Ksp = [Zn2+]^2.[Fe(CN)6^4-]

with,

[Zn2+] = 0.13 M

we get,

[Fe(CN)6^4-] = 2.1 x 10^-16/(0.13)^2

                      = 1.24 x 10^-14 M

(b) [Fe(CN)6^4-] = Ksp/[Zn2+]^2

with,

[Zn2+] = 4.9 x 10^-7 M

we get,

concentration of K4Fe(CN)6 needed = 2.1 x 10^-16/(4.9 x 10^-7)^2

                                                           = 8.75 x 10^-4 M

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