A solution which remains in equilibrium with undissolved solute is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product `(K_(sp))`. For the electrolyte, `A_(x),B_(y),:A_(x),B_(y(s)) rarr xA^(y+)+ y^(Bx-)`, with solubility S, the solubility product `(K_(sp)) =x^(x)xxy^(y) xx s^(x+y)`. While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an clectrolyte exceeds its `K_(sp)`, value at a particular temperature, then precipitation occurs.
The solubility of `BaSO_(4)`, in 0.1 M `BaCl_(2)`, solution is `(K_(sp)`, of `BaSO_(4), = 1.5 xx 10^(-9))`

Options

A. a) `1.5xx10^(-9)M`
B. b) `1.5xx10^(-8)` M
C. c) `2.25xx10^(-16)M`
D. d) `2.25xx10^(-18) M`

Answer

To solve the problem of finding the solubility of \( \text{BaSO}_4 \) in a \( 0.1 \, \text{M} \, \text{BaCl}_2 \) solution, we will follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( \text{BaSO}_4 \) in water can be represented as:
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

### Step 2: Define solubility and concentrations
Let the solubility of \( \text{BaSO}_4 \) in the solution be \( S \). In a saturated solution of \( \text{BaSO}_4 \), the concentration of \( \text{Ba}^{2+} \) ions will be \( S \) and the concentration of \( \text{SO}_4^{2-} \) ions will also be \( S \).

However, since we are dissolving \( \text{BaSO}_4 \) in a \( 0.1 \, \text{M} \, \text{BaCl}_2 \) solution, the concentration of \( \text{Ba}^{2+} \) ions from \( \text{BaCl}_2 \) will also contribute to the total concentration of \( \text{Ba}^{2+} \) ions in the solution. Therefore, the total concentration of \( \text{Ba}^{2+} \) ions will be:
\[ [\text{Ba}^{2+}] = S + 0.1 \]
And the concentration of \( \text{SO}_4^{2-} \) ions will be:
\[ [\text{SO}_4^{2-}] = S \]

### Step 3: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) for \( \text{BaSO}_4 \) is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]
Substituting the concentrations we have:
\[ K_{sp} = (S + 0.1)(S) \]

### Step 4: Substitute the known value of \( K_{sp} \)
Given that \( K_{sp} = 1.5 \times 10^{-9} \), we can write:
\[ (S + 0.1)S = 1.5 \times 10^{-9} \]

### Step 5: Assume \( S \) is much smaller than \( 0.1 \)
Since \( \text{BaSO}_4 \) is sparingly soluble, we can assume \( S \) is much smaller than \( 0.1 \). Thus, we can approximate:
\[ S + 0.1 \approx 0.1 \]
This simplifies our equation to:
\[ 0.1S = 1.5 \times 10^{-9} \]

### Step 6: Solve for \( S \)
Now, we can solve for \( S \):
\[ S = \frac{1.5 \times 10^{-9}}{0.1} = 1.5 \times 10^{-8} \]

### Conclusion
The solubility of \( \text{BaSO}_4 \) in \( 0.1 \, \text{M} \, \text{BaCl}_2 \) solution is:
\[ S = 1.5 \times 10^{-8} \, \text{M} \]