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The solubility of manganese carbonate MnCO₃ is 4.8x10^-4 g/l. what is the solubility oproduct constant for manganese carbonate?

User John Guan
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1 Answer

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Final answer:

The solubility product constant (Ksp) of manganese carbonate (MnCO₃) can be calculated by first converting the solubility from grams per liter to molarity using the molar mass, and then squaring the molar solubility.

Step-by-step explanation:

To find the solubility product constant (Ksp) for manganese carbonate (MnCO₃), we need to consider the dissolution reaction that occurs in water: MnCO₃(s) → Mn²⁺(aq) + CO₃²⁺(aq)

When MnCO₃ dissolves, it produces one mole of Mn²⁺ and one mole of CO₃²⁺ for each mole of MnCO₃ that dissolves. The solubility given as 4.8x10^-4 g/L needs to be converted to molarity (M) by using the molar mass of MnCO₃.

The molecular weight of MnCO₃ is approximately 114.95 g/mol. Hence, the molar solubility (s) can be calculated by

s = (4.8x10^-4 g/L) / (114.95 g/mol) Once the molar solubility is calculated, the Ksp can be determined since: Ksp = [Mn²⁺] [CO₃²⁺] = s * s = s²

The Ksp will therefore be the molar solubility squared.

User Deepesh Uniyal
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