ANALYTICAL QUESTION HELP

Question

Determine the molar solubility of copper(I) azide (CuN3CuN3) in a solution with a pH of 10.20. Ignore activities. The Ksp for CuN3 is 4.9×10−9. The Ka for HN3 is 2.2×10−5

Answer 1

(1). CuN₃(s) <==> Cu⁺(aq) + N₃^-(aq) ... Ksp = 4.9x10^-9

(2). HN₃(aq) <==> H⁺(aq) + N₃^-(aq) ... Ka = 2.2x10^-5

(3). Reversing eq.2 we have H⁺(aq) + N₃^-(aq) <==> HN₃(aq) ... 1/Ka = 4.55x10⁴

Adding (1) and (3) we have ...

CuN₃(s) <==> Cu⁺(aq) + N₃^-(aq) ... Ksp = 4.9x10^-9

H⁺(aq) + N₃^-(aq) <==> HN₃(aq) ... 1/Ka = 4.55x10⁴

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CuN₃(s) + H⁺(aq) <==> Cu⁺(aq) + HN₃(aq) ... Keq = 4.9x10^-9 x 4.55x10⁴ = 2.23x10^-4

From pH = 10.2, [H⁺] = 1x10-^10.2 = 6.31x10^-11 M

2.23x10^-4 = [Cu⁺][HN₃] / [H⁺]

2.23x10^-4 = (x)(x) / 6.31x10^-11

x² = 1.41x10^-14

x = 1.19x10^-7 M = solubility of CuN₃

(as usual, be sure to check all of the math)

Answer 2

It seems to me that the molar solubility of CuN₃ is unaffected by the high pH solution. x₁, the extent of dissociation of CuN₃ in water would be sqrt(4.9 x 10^-9). If x₂ is the extent of the reaction of N₃^- reacting with H⁺ to form HN₃, it must be less than the concentration of H⁺ = 10^-10.2 = 6.31x 10^-16 which is negligible compared to x₁. Sorry if I misunderstood the problem.