To find the solubility of magnesium carbonate (MgCO₃) in water, we use the solubility product constant (Ksp). The Ksp for MgCO₃ is given as $3.5 \times 10^{-8}$.

The dissolution of MgCO₃ in water is represented by the following equilibrium reaction:

$\text{MgCO}_3 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + \text{CO}_3^{2-} (aq)$

Let the solubility of MgCO₃ in mol/L be $s$. This means that at equilibrium, the concentration of $\text{Mg}^{2+}$ ions and $\text{CO}_3^{2-}$ ions will both be equal to $s$, since for every mole of MgCO₃ that dissolves, one mole of each ion is produced.

The Ksp expression is:

[ K_{sp} = [\text{Mg}^{2+}][\text{CO}_3^{2-}] ]

Since both concentrations are equal to $s$, we can substitute them into the expression:

$K_{sp} = s \times s = s^2$

Now, solving for $s$:

$3.5 \times 10^{-8} = s^2$

$s = \sqrt{3.5 \times 10^{-8}}$

$s = 5.92 \times 10^{-4} \, \text{mol/L}$

Thus, the solubility of MgCO₃ in water is approximately $5.92 \times 10^{-4} \, \text{mol/L}$.

Would you like further details or have any questions?

Here are 5 related questions to further explore this topic:

1. How does the solubility of MgCO₃ change in the presence of common ions?
2. What effect does temperature have on the solubility of MgCO₃?
3. How would you calculate the solubility of MgCO₃ in a solution with pH = 7?
4. How is the solubility product related to the common ion effect?
5. Can you calculate the molar mass of MgCO₃ and convert the solubility into grams per liter?

Tip: When calculating solubility, always double-check your Ksp expression to ensure you're accounting for the correct ion concentrations.