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 product at that temperature is

Options

A. `10^(-4)M^(2)`
B. `9.18xx10^(-4) M^(2)`
C. `10^(-6) M^(2)`
D. `9.18xx10^(-8)M^(2)`

Answer

To find the solubility product (Ksp) of lead sulfate (PbSO4) given its solubility in water, we can follow these steps:

### Step 1: Understand the solubility of PbSO4
The solubility of PbSO4 in water is given as 0.303 g/L at 25°C.

### Step 2: Calculate the molar mass of PbSO4
To convert solubility from grams per liter to moles per liter, we first need to calculate the molar mass of PbSO4.
- Molar mass of Pb = 207 g/mol
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol × 4 = 64 g/mol

Adding these together:
\[ \text{Molar mass of PbSO4} = 207 + 32 + 64 = 303 \text{ g/mol} \]

### Step 3: Convert solubility from g/L to mol/L
Now, we can convert the solubility from grams per liter to moles per liter:
\[ \text{Solubility (S)} = \frac{0.303 \text{ g/L}}{303 \text{ g/mol}} = 0.001 \text{ mol/L} = 10^{-3} \text{ mol/L} \]

### Step 4: Write the dissociation equation for PbSO4
The dissociation of lead sulfate in water can be represented as:
\[ \text{PbSO}_4 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

### Step 5: Determine the concentrations of ions at equilibrium
From the dissociation equation, we see that for every mole of PbSO4 that dissolves, one mole of Pb²⁺ and one mole of SO₄²⁻ are produced. Therefore, at equilibrium:
- \[ [\text{Pb}^{2+}] = S = 10^{-3} \text{ mol/L} \]
- \[ [\text{SO}_4^{2-}] = S = 10^{-3} \text{ mol/L} \]

### Step 6: Calculate the solubility product (Ksp)
The solubility product (Ksp) is given by the expression:
\[ K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}] \]
Substituting the values:
\[ K_{sp} = (10^{-3})(10^{-3}) = 10^{-6} \]

### Final Answer
The solubility product (Ksp) of PbSO4 at 25°C is:
\[ K_{sp} = 10^{-6} \]

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