Calcium carbonate, CaCO3, has a solubility in water of 0.0180 g/l at 25 C. Calculate the solubility product (Ksp) for CaCO3
The correct answer and explanation is :
To calculate the solubility product constant (Ksp) for calcium carbonate (CaCO₃), follow these steps:
Step 1: Write the Dissociation Equation
When calcium carbonate dissolves in water, it dissociates as:
[
CaCO_3 (s) \rightleftharpoons Ca^{2+} (aq) + CO_3^{2-} (aq)
]
From this equation, 1 mole of CaCO₃ produces 1 mole of Ca²⁺ and 1 mole of CO₃²⁻.
Step 2: Convert Solubility to Molarity
The given solubility of CaCO₃ is 0.0180 g/L at 25°C. To find the molar solubility, divide by its molar mass:
[
\text{Molar Mass of CaCO₃} = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \, g/mol
]
[
\text{Molar Solubility} = \frac{0.0180 \, g/L}{100.09 \, g/mol} = 1.80 \times 10^{-4} \, M
]
Since CaCO₃ dissociates in a 1:1 ratio, the equilibrium concentrations of ions are:
[
[Ca^{2+}] = 1.80 \times 10^{-4} \, M
]
[
[CO_3^{2-}] = 1.80 \times 10^{-4} \, M
]
Step 3: Calculate Ksp
The solubility product expression for CaCO₃ is:
[
K_{sp} = [Ca^{2+}][CO_3^{2-}]
]
Substituting the values:
[
K_{sp} = (1.80 \times 10^{-4}) \times (1.80 \times 10^{-4})
]
[
K_{sp} = 3.24 \times 10^{-8}
]
Final Answer:
[
K_{sp} = 3.24 \times 10^{-8}
]
Explanation (300 Words)
The solubility product constant (Ksp) is a measure of the extent to which a sparingly soluble compound dissolves in water. In this case, CaCO₃ dissociates into calcium ions (Ca²⁺) and carbonate ions (CO₃²⁻) in a 1:1 ratio.
We first determined the molar solubility of calcium carbonate by converting the given solubility in grams per liter to moles per liter using the molar mass. The calculated molar solubility was 1.80 × 10⁻⁴ M, which represents the concentration of both Ca²⁺ and CO₃²⁻ in solution.
Using the solubility product expression Ksp = [Ca²⁺][CO₃²⁻], we substituted the molar solubility values into the equation. Since both ion concentrations are equal, we squared the molar solubility to obtain Ksp = 3.24 × 10⁻⁸.
This Ksp value is a fundamental property of CaCO₃ and is useful for predicting precipitation reactions. For example, if the product of Ca²⁺ and CO₃²⁻ concentrations in a solution exceeds Ksp, precipitation will occur. Conversely, if the product is below Ksp, more CaCO₃ can dissolve.
In practical applications, this principle is crucial in geology (limestone dissolution), water treatment (scale formation), and biology (bone formation and shell formation in marine organisms). Additionally, changes in pH affect carbonate solubility since CO₃²⁻ reacts with H⁺ to form bicarbonate (HCO₃⁻), increasing CaCO₃ solubility in acidic conditions.
Thus, the solubility product constant quantifies the limited solubility of CaCO₃ and helps in predicting its behavior in different chemical and environmental conditions.