If the K_sp for zinc carbonate (ZnCO₃) is 1.46×10^-10 at $25 ^ { \circ } \mathrm { C }$ , what is its aqueous solubility in terms of the molarity of ions produced in solution?

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The solubility product (Ksp) for zinc carbonate (ZnCO3) is given as 1.46×10^-10. The dissolution of zinc carbonate in water can be represented by the equation: ZnCO3(s) ⇌ Zn^2+(aq) + CO3^2-(aq). From this equation, it's clear that for every mole of ZnCO3 that dissolves, 1 mole of Zn^2+ and 1 mole of CO3^2- are produced. Therefore, if s is the solubility of ZnCO3 in mol/L, then the Ksp expression is: Ksp = [Zn^2+][CO3^2-] = s * s = s^2. Solving for s gives: s = sqrt(Ksp) = sqrt(1.46×10^-10) ≈ 1.21×10^-5 M.

If the Ksp for Zinc Carbonate (ZnCO3) Is 1 25∘C25 ^ { \circ } \mathrm { C }25∘C

If the K_sp for Zinc Carbonate (ZnCO₃) Is 1 $25 ^ { \circ } \mathrm { C }$