Question

Ignoring activities, determine the molar solubility of copper (I) azide (CuN3) in a solution with a pH of 3.088. Ksp (CuN3) = 4.9

Answer 1

CuN3 Solution, pH = 3.088

Ksp of CuN3 = 4.9 * 10^-9 & Ka of HN3 = 2.2 * 10^-5

P^H = 3.088

We knw that , pH = -Log [H⁺]

                       [H⁺] = 10^-PH = 10^-3.088

                                 [H⁺] = 8.16 * 10^-4

HN3 H⁺ + N3^-

We know that, Ka = [H⁺ ][N3^-] / [HN3] = 2.2 * 10^-5

                                   Ksp = [Cu⁺][N3^-] = 4.9 *10^-9

                                  Keq = Ksp * 1 / Ka

                       Keq = [Cu⁺][N3^-] * [HN3] / [H⁺ ][N3^-]

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

                    Keq    = 4.9 * 10^-9 / 2.2 * 10^-5

                             Keq      = 2.23 * 10^-14

here y = Molar Solubility of CuN3

      Y = [Cu⁺]=[HN3]

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

2.23 * 10^-4 = Y² / 8.16 * 10^-4

               Y² = 2.23 * 10^-4 * 8.16 * 10^-4

            Y²      = 1.82 * 10^-7

Molar Solubility of CuN3,    Y       = 4.3 * 10^-4