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Ignoring activities, determine the molar solubility of copper (I) azide (CuN3) in a solution with a pH of 10.83. Ksp (CuN3) = 4.9 × 10–9; Ka (HN3) = 2.2 × 10–5.

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    pH = 10.83; (H^+) = 1.48E-11
    CuN3 ==> Cu^+ + N3^-
    H^+ + N3^- ==> HN3 add the two eqns.
    ------------------- N3^- cancels.
    CuN3 + H^+ ==> Cu^+ + HN3
    Keq for this rxn is Ksp/Ka = 0.000223
    ..x....1.48E-11..x.....x

    Keq = 0.000223 = (x)(x)/1.48E-11
    Solve for x.

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