Question

Calculate the molar solubility of magnesium phosphate, Mg3(PO4)2 (Ksp = 1.0×10–25)


Calculate the concentration of lithium ions in a saturated solution of lithium phosphate, Li3PO4 (Ksp = 3.2×10–9)

Answer 1

    ^{given k}_sp=1.0*10^-25

Mg₃(PO₄)₂          3Mg²⁺   + 2PO₄^3-

    let molar solubility be X

                            (3X)3 for magnesium ion

                           (2X)² for phosphate ion

       k_sp=(3X)³(2X)²

         1*(10)^-25=(3X)³(2X)²

          10^-25=27*4(X)⁵

         by doing calculations we get molar solubility=3.92*(10)^-6 mol/L

b)given k_sp=3.2*(10)^-

         Li₃PO₄            3Li⁺ + PO₄^3-

           ^{Let,concentration be X}

          ^Ksp=(3X)³(X)

     3.2*(10)^-9 =27(X)⁴

      X =3.3*(10)^-3Mol/L

    concentration of lithium ions are 3X

thersfore ,

concentration=3*3.3(10)^-3=9.9*(10)^-3