For caco3 ksp 8.0 x 10-13 at 25c what is the solubility of caco3 at this temperature?
To determine the solubility of \( \text{CaCO}_3 \) at 25°C using its \( K_{sp} \) (solubility product constant), you need to set up an equation and solve for the concentration of calcium ions (\( \text{Ca}^{2+} \)) and carbonate ions (\( \text{CO}_3^{2-} \)) in the solution.
The solubility product constant (\( K_{sp} \)) is an equilibrium constant that describes the degree to which a compound can dissolve in water. For a sparingly soluble salt like \( \text{CaCO}_3 \), its solubility product can be written as:
\( K_{sp} = [\text{Ca}^{2+}][\text{CO}_3^{2-}] \)
Where:
\([ \text{Ca}^{2+} ]\) represents the concentration of calcium ions in the solution,
\([ \text{CO}_3^{2-} ]\) represents the concentration of carbonate ions in the solution.
The given \( K_{sp} \) for \( \text{CaCO}_3 \) at 25°C is \( 8.0 \times 10^{-13} \). Since \( \text{CaCO}_3 \) dissociates into one calcium ion and one carbonate ion, we can assume that the concentration of \( \text{Ca}^{2+} \) and \( \text{CO}_3^{2-} \) is the same in the solution.
Let's define \( x \) as the concentration of \( \text{Ca}^{2+} \) or \( \text{CO}_3^{2-} \) in moles per liter (M):
\( K_{sp} = x \times x \)
Now, substitute the value of \( K_{sp} \) and solve for \( x \):
\( 8.0 \times 10^{-13} = x^2 \)
To solve for \( x \), take the square root of both sides:
\( x = \sqrt{8.0 \times 10^{-13}} \)
Now, calculate the value of \( x \):
\( x = 8.9 \times 10^{-7} \, \text{M} \)
Therefore, the solubility of \( \text{CaCO}_3 \) at 25°C is approximately \( 8.9 \times 10^{-7} \, \text{M} \).