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 `PbSO_(4)`, in water is 0.303 g/l at `25^(@)`C, its solubility at that temperature is

Options

A. `1.8xx10^(-5)M`
B. `(4.8)xx10^(-5)M`
C. `sqrt(3)xx10^(-6)M`
D. `10^(-6)`M

Answer

To find the solubility of PbSO4 in water at 25°C, we can follow these steps:

### Step 1: Understand the Dissociation of PbSO4
When PbSO4 dissolves in water, it dissociates into lead ions (Pb²⁺) and sulfate ions (SO4²⁻):
\[ \text{PbSO}_4 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

### Step 2: Define Solubility (S)
Let the solubility of PbSO4 be denoted as \( S \). Therefore, at equilibrium:
- The concentration of Pb²⁺ ions = \( S \)
- The concentration of SO4²⁻ ions = \( S \)

### Step 3: Write the Expression for Solubility Product (Ksp)
The solubility product \( K_{sp} \) for the dissociation of PbSO4 can be expressed as:
\[ K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}] \]
Substituting the concentrations gives:
\[ K_{sp} = S \times S = S^2 \]

### Step 4: Calculate the Solubility in Moles per Liter
Given that the solubility of PbSO4 in water is 0.303 g/L, we need to convert this to moles per liter. First, we calculate the molar mass of PbSO4:
- Molar mass of Pb = 207.2 g/mol
- Molar mass of S = 32.07 g/mol
- Molar mass of O = 16.00 g/mol × 4 = 64.00 g/mol
- Total molar mass of PbSO4 = 207.2 + 32.07 + 64.00 = 303.27 g/mol

Now, we can calculate the solubility in moles per liter:
\[ S = \frac{0.303 \text{ g/L}}{303.27 \text{ g/mol}} \approx 0.000999 \text{ mol/L} \]
This can be approximated to:
\[ S \approx 1.0 \times 10^{-3} \text{ mol/L} \]

### Step 5: Calculate Ksp
Now we can calculate \( K_{sp} \):
\[ K_{sp} = S^2 = (1.0 \times 10^{-3})^2 = 1.0 \times 10^{-6} \text{ mol}^2/\text{L}^2 \]

### Step 6: Conclusion
Thus, the solubility product \( K_{sp} \) of PbSO4 at 25°C is:
\[ K_{sp} = 1.0 \times 10^{-6} \text{ mol}^2/\text{L}^2 \]