Consider a solution that is 2.0×10−2 m in fe2+ and 1.8×10−2 m in mg2+. (ksp for feco3 is 3.07×10−11 and ksp for mgco3 is 6.82×10−6.)

-what minimum concentration of k2co3 is required to cause the precipitation of the cation that precipitates first?
express your answer using two significant figures.

The minimum concentration of K₂CO₃ is (1.5 x 10⁻⁹ M) that is required to cause the the precipitation of FeCO₃ firstly.

Explanation:

FeCO₃ is formed when: [Fe²⁺][CO₃⁻²] > Ksp of FeCO₃.

MgCO₃ is formed when: [Mg²⁺][CO₃⁻²] > Ksp of MgCO₃.

1) For FeCO₃:

∵ Ksp (FeCO₃) = [Fe²⁺][CO₃⁻²]

(3.07 × 10⁻¹¹) = (2.0 × 10²⁻)[CO₃⁻²]

∴ [CO₃⁻²] = (3.07 × 10⁻¹¹)/(2.0 × 10²⁻) = 1.5 x 10⁻⁹ M.

2) For MgCO₃:

∵ Ksp (MgCO₃) = [Mg²⁺][CO₃⁻²]

(6.82 × 10⁻⁶) = (1.8 × 10²⁻)[CO₃⁻²]

∴ [CO₃⁻²] = (6.82 × 10⁻⁶)/(1.8 × 10²⁻) = 3.8 x 10⁻⁴ M.

The concentration of K₂CO₃ is (1.5 x 10⁻⁹ M) that is required to cause the the precipitation of FeCO₃ is much less than the concentration of K₂CO₃ (3.8 x 10⁻⁴ M) that is required to cause the the precipitation of MgCO₃.

So, The minimum concentration of K₂CO₃ is (1.5 x 10⁻⁹ M) that is required to cause the the precipitation of FeCO₃ firstly.

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.

it’s previous models   because when i picked that son of a gun’s answer it was wrong look for previous models because they might make it a different letter