Find the solubility of `Ca(IO_(3))_(2)` is `molL^(-1)` in a solution containing `0.1M CaC1` at `25^(@)C`. `K_(sp)` of `Ca(IO_(3))_(2) = 6.3 xx 10^(-7)`

Explanation

To find the solubility of \( \text{Ca(IO}_3)_2 \) in a solution containing \( 0.1 \, \text{M} \, \text{CaCl}_2 \) at \( 25^\circ C \), we can follow these steps:

### Step 1: Write the dissociation equation for \( \text{Ca(IO}_3)_2 \)
The dissociation of calcium iodate can be represented as:
\[ \text{Ca(IO}_3)_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{IO}_3^{-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( \text{Ca(IO}_3)_2 \) in the solution be \( s \, \text{mol/L} \). Therefore, at equilibrium:
- The concentration of \( \text{Ca}^{2+} \) ions will be \( 0.1 + s \)
- The concentration of \( \text{IO}_3^{-} \) ions will be \( 2s \)

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for \( \text{Ca(IO}_3)_2 \) is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{IO}_3^{-}]^2 \]
Substituting the equilibrium concentrations:
\[ K_{sp} = (0.1 + s)(2s)^2 \]

### Step 4: Substitute \( K_{sp} \) value
Given \( K_{sp} = 6.3 \times 10^{-7} \), we can substitute this into the equation:
\[ 6.3 \times 10^{-7} = (0.1 + s)(4s^2) \]

### Step 5: Make an assumption
Since \( s \) is expected to be very small compared to \( 0.1 \, \text{M} \), we can approximate:
\[ 0.1 + s \approx 0.1 \]
Thus, the equation simplifies to:
\[ 6.3 \times 10^{-7} = 0.1 \times 4s^2 \]

### Step 6: Solve for \( s \)
Rearranging the equation gives:
\[ s^2 = \frac{6.3 \times 10^{-7}}{0.4} \]
Calculating this:
\[ s^2 = 1.575 \times 10^{-6} \]
Taking the square root:
\[ s = \sqrt{1.575 \times 10^{-6}} \approx 0.00125 \, \text{mol/L} \]

### Final Answer
The solubility of \( \text{Ca(IO}_3)_2 \) in the solution is approximately:
\[ s \approx 1.25 \times 10^{-3} \, \text{mol/L} \]

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