Question

Can someone help me with chemistry, Answer below is wrong!

gnoring activities, determine the molar solubility of copper (1) azide (CuN3) in a solution with a pH of 3.960. Ksp (CuNs) = 4.9 10-9, Ka (HN3)-2241 0-5 Number 4.45 x 10-5

Answer 1

pH = 3.96;

Hence, [H+] = 1.096 x 10^-4

HN3 --> H+ + N3-

Ka = [H+][N3-]/ [HN3]

= 2.2 x 10^-5

Ksp = [Cu+][N3-]

= 4.9 x 10^-9

CuN3(s) --> Cu+ + N3-

H+ + N3- --> HN3

Add the above two equations:

  

CuN3(s) + H+ --> Cu+ + HN3

Keq = [Cu+][HN3] / [H+]

Rearrange Ka to have [HN3] / [H+] [N3-]

Ka = [H+][N3-] / [HN3]

[HN3] / [H+][N3-] = 1/Ka

Keq = Ksp x 1/Ka = [Cu+][N3-] x [HN3] /[H+][N3-]

= [Cu+][HN3] / [H+]

Keq = 4.9 x 10^-9 / 2.2 x 10^-5 = 2.23 x 10^-4

Let Y = molar solubility of CuN3

Y = [Cu+] = [HN3]

Keq = 2.23 x 10^-4 = Y(Y) / [H+]

2.23 x 10^-4 = (Y)(Y) / 1.096 x 10^-4

Y^2 = 2.44 x 10^-8

Y = 1.56 x 10^-4

Hence molar solubility of CuN3 = 1.56 x 10^-4