Question

Excess solid Na₂CO₃ is added to a solution containing 0.400 M (each) Mg²⁺ and Zn²⁺ ions. Ksp for MgCO₃ is 3.50 × 10⁻⁸ and Ksp for ZnCO₃ is 1.00 × 10⁻¹⁰. ZnCO₃, with the smaller Ksp, will be the least soluble and will begin precipitating first. What will be the [Zn²⁺] concentration when MgCO₃ just begins to precipitate? (Assume no volume change upon addition of the solid Na₂CO₃).

Answer 1

To solve this, we need Ksp values. Ksp of MgCO3= [Mg+2][CO2] = 3.50 x 10-8 Ksp of ZnCO3= [Zn+2][CO, 2] = 1.00 x 10-10 Given that Concentration of Mg+2 = 0.400 M Concentration of Zn +2 = 0.400 M. First calculate the minimum [CO3-2) ions for the precipitation of both. Ksp 3.50 x10-8 — = 8.75x10-8 M [CO3 min. for Mgco - - Mg +27 0.400 Ksp [CO3-2 Imin.for Znco> [Fe21 1.00x10-10 -= 2.50x10-10M 0.400 Since [CO3:2] for ZnCoz is lower than MgCO3, ZnCO3 will precipitate first. Now, if C032 ions are continuously added, at some time MgCO3 will be precipitated out. [CO3:2] required for the precipitation of MgCO3 is 8.75 10-8 M That is when the conc. of Coz 2 is 8.75 x 10-8 M, MgCO3 starts to precipitate out. To find the [Zn+2] in the solution when MgCO, start to precipitate out. ([CO-2)= 8.75 x 10-8 M in the solution.) -21 Zn+2]=_^sp 014 Kep of ZnCO; [CO; ? ]min.for Meco. 1.00×10-10 8.75x10-8 -e = 1.14x10-M