What is the solubility of magnesium carbonate (MgCO3) in water at 25 °C? Given the
data for MgCO3: molar mass = 84 g mol–1 and Ksp at 25 °C = 6.8 x 10
–6
The solubility of magnesium carbonate (MgCO3) in water at 25 °C can be determined using the solubility product constant (Ksp).
The Ksp for MgCO3 at 25 °C is given as 6.8 x 10^-6.
The equation for the dissolution of MgCO3 in water is as follows:
MgCO3(s) ↔ Mg2+(aq) + CO3^2-(aq)
From this equation, we can see that 1 mole of MgCO3 will dissolve to form 1 mole of Mg2+ and 1 mole of CO3^2-.
Since the molar mass of MgCO3 is given as 84 g/mol, we can calculate the solubility of MgCO3 in water as follows:
1 mole of MgCO3 = 84 g
1 mole of Mg2+ = 24 g (atomic mass of Mg = 24 g/mol)
1 mole of CO3^2- = 60 g (atomic mass of C = 12 g/mol, atomic mass of O = 16 g/mol, so CO3^2- has a molar mass of 60 g/mol)
Using the Ksp expression for MgCO3, we have:
Ksp = [Mg2+][CO3^2-]
Let's assume the solubility of MgCO3 in water is x mol/L.
Thus, the concentration of Mg2+ and CO3^2- in the saturated solution will also be x mol/L.
Substituting these values into the Ksp expression:
6.8 x 10^-6 = (x)(x)
Simplifying:
6.8 x 10^-6 = x^2
Taking the square root of both sides:
x = √(6.8 x 10^-6)
x ≈ 0.00261 mol/L
Therefore, the solubility of magnesium carbonate (MgCO3) in water at 25 °C is approximately 0.00261 mol/L.