Answer:
The minimum concentration of K₂CO₃ is (1.5 x 10⁻⁹ M) that is required to cause the the precipitation of FeCO₃ firstly.
Explanation:
- Firstly, we should know that the precipitate is formed when the ionic product of the ions is higher than the solubility product.
FeCO₃ is formed when: [Fe²⁺][CO₃⁻²] > Ksp of FeCO₃.
MgCO₃ is formed when: [Mg²⁺][CO₃⁻²] > Ksp of MgCO₃.
<u><em>1) For FeCO₃:</em></u>
∵ 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.
<u><em>2) For MgCO₃:</em></u>
∵ 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.
<em>So, The minimum concentration of K₂CO₃ is (1.5 x 10⁻⁹ M) that is required to cause the the precipitation of FeCO₃ firstly.</em>
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Group I and Group II have the same number of outermost electrons as you go down each group but the shells increase therefore as you go downward it becomes much easier to remove electrons because of its wide radius however group 7 and 6 have seven and six electrons in their outermost shell respectively. Therefore down the group it is much easier to attract electrons and across the period it is much harder because the number of electrons on the outermost shell increase