Question

Determine the molar solubility of copper(1) azide (CuN,) in a solution with a pH of 8.59. Ignore activities. The Ksp for CuN, is 4.9 x 10-9. The K, for HN, is 2.2 x 10-5 molar solubility:

Answer 1

Na HN - U+ + N₂ ph = 8.59 pn = -log [n+ 10-8.59 = Intl - 2.57 x 10 m ka = [n+][N3] [UN31 2.2 x 105 = [ut] {N} Tunal } - ② ③ CuNa → Cut + N3- Ksp = [Cu+] [N31 4.9x109 = [Cut] [ N3] Adding eq 2 ③ Cu Na ut +N3 M + + Ng - HN₃ cu Nz dnt wit of uNz keg = [cut Jenna [nt] + ka = [ut] (N3") Гим; 1 I'UNI [ut] [N ] Ko keq = kep XL Ka

-) [Cu+I (N3] x (uN3] l'n + ] [N ] - [Cut] una [ut] - keq = 4.9x10 9 = 2.23x10 4. 2.2 8 10 5 let sa molas solubility of cu N3 s rout] - [UN3 ] keq = 2.23x10 4 = S(ol (nt] - 2.23 X 10 = ² [2:57X10**] ~ 52 = 5.73x10-13 7.57 x 10-7M. Answer