If solutbility of `Ca(IO_(3))_(2)` in water at `20^(@)C` is `3.9gL^(-1)`. Calculate the `K_(sp)`. Given `Mw Ca(IO_(3))_(2) = 390`.

Explanation

To calculate the solubility product constant (Ksp) for calcium iodate, \( Ca(IO_3)_2 \), given its solubility in water at \( 20^\circ C \) is \( 3.9 \, g/L \) and the molecular weight is \( 390 \, g/mol \), we can follow these steps:

### Step 1: Convert solubility from grams per liter to moles per liter
To find the molarity (moles per liter) of the solute, we use the formula:

\[ \text{Molarity (M)} = \frac{\text{mass (g)}}{\text{molecular weight (g/mol)}} \]

Substituting the values:

\[ \text{Molarity (M)} = \frac{3.9 \, g}{390 \, g/mol} = 0.01 \, mol/L \]

### Step 2: Write the dissociation equation for \( Ca(IO_3)_2 \)
When \( Ca(IO_3)_2 \) dissolves in water, it dissociates according to the following equation:

\[ Ca(IO_3)_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2 IO_3^{-} (aq) \]

### Step 3: Define the solubility in terms of \( S \)
Let \( S \) be the solubility of \( Ca(IO_3)_2 \) in moles per liter. From the dissociation equation, we can see that:

- The concentration of \( Ca^{2+} \) ions will be \( S \).
- The concentration of \( IO_3^{-} \) ions will be \( 2S \).

### Step 4: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) is given by the expression:

\[ K_{sp} = [Ca^{2+}][IO_3^{-}]^2 \]

Substituting the concentrations in terms of \( S \):

\[ K_{sp} = [S][2S]^2 = S \cdot 4S^2 = 4S^3 \]

### Step 5: Substitute the value of \( S \)
Now we substitute \( S = 0.01 \, mol/L \) into the \( K_{sp} \) expression:

\[ K_{sp} = 4(0.01)^3 = 4 \times 10^{-6} \]

### Final Answer
Thus, the solubility product constant \( K_{sp} \) for \( Ca(IO_3)_2 \) is:

\[ K_{sp} = 4 \times 10^{-6} \]

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