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.

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.