The solubility product of chalk is `9.3xx10^(-8)`. Calculate its solubility in gram per litre

Options

A. `0.3040 gram//litre`
B. `0.0304 gram//litre`
C. `2.0304 gram//litre`
D. `4.0304 gram//litre`

Explanation

To calculate the solubility of chalk (calcium carbonate, CaCO₃) in grams per liter given its solubility product (Ksp), we can follow these steps:

### Step 1: Write the dissociation equation
Chalk (CaCO₃) dissociates in water as follows:
\[ \text{CaCO}_3 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{CO}_3^{2-} (aq) \]

### Step 2: Define solubility (S)
Let the solubility of CaCO₃ be \( S \) mol/L. When 1 mole of CaCO₃ dissolves, it produces 1 mole of \( \text{Ca}^{2+} \) ions and 1 mole of \( \text{CO}_3^{2-} \) ions. Therefore:
- Concentration of \( \text{Ca}^{2+} \) = \( S \)
- Concentration of \( \text{CO}_3^{2-} \) = \( S \)

### Step 3: Write the expression for Ksp
The solubility product (Ksp) is given by:
\[ Ksp = [\text{Ca}^{2+}][\text{CO}_3^{2-}] = S \times S = S^2 \]

### Step 4: Substitute the given Ksp value
We are given that:
\[ Ksp = 9.3 \times 10^{-8} \]
Thus:
\[ S^2 = 9.3 \times 10^{-8} \]

### Step 5: Solve for S
To find \( S \), take the square root of both sides:
\[ S = \sqrt{9.3 \times 10^{-8}} \]
Calculating this gives:
\[ S \approx 3.04 \times 10^{-4} \text{ mol/L} \]

### Step 6: Convert moles to grams
To convert the solubility from moles per liter to grams per liter, we need the molar mass of calcium carbonate (CaCO₃). The molar mass is calculated as follows:
- Molar mass of Ca = 40 g/mol
- Molar mass of C = 12 g/mol
- Molar mass of O (3 atoms) = 16 g/mol × 3 = 48 g/mol

Total molar mass of CaCO₃:
\[ \text{Molar mass of CaCO}_3 = 40 + 12 + 48 = 100 \text{ g/mol} \]

Now, convert the solubility in moles to grams:
\[ \text{Solubility in g/L} = S \times \text{Molar mass} \]
\[ \text{Solubility in g/L} = 3.04 \times 10^{-4} \text{ mol/L} \times 100 \text{ g/mol} \]
\[ \text{Solubility in g/L} = 3.04 \times 10^{-2} \text{ g/L} \]

### Step 7: Final answer
Thus, the solubility of chalk in grams per liter is:
\[ \text{Solubility} \approx 0.0304 \text{ g/L} \]