The solubility product of `BaSO_4 ` is `4 xx 10^(-10)` . The solubility of `BaSO_4` in presence of 0.02 `N H_2SO_4` will be

Options

A. `4 xx 10^(-8)` M
B. `2 xx 10^(-8)` M
C. `2 xx 10^(-5) `M
D. `2 xx 10^(-4)` M

Answer

To find the solubility of barium sulfate (BaSO₄) in the presence of 0.02 N H₂SO₄, we will use the concept of solubility product (Ksp) and the common ion effect. Here’s a step-by-step solution:

### Step 1: Write the dissociation equations
Barium sulfate dissociates in water as follows:
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

Sulfuric acid (H₂SO₄) dissociates completely in water:
\[ \text{H}_2\text{SO}_4 (aq) \rightarrow 2\text{H}^+ (aq) + \text{SO}_4^{2-} (aq) \]

### Step 2: Define the solubility (S)
Let the solubility of BaSO₄ in the presence of H₂SO₄ be \( S \). Thus, at equilibrium:
- The concentration of \(\text{Ba}^{2+}\) ions will be \( S \).
- The concentration of \(\text{SO}_4^{2-}\) ions will be \( S + 0.02 \) (because of the contribution from H₂SO₄).

### Step 3: Write the expression for Ksp
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]
Substituting the concentrations:
\[ K_{sp} = S \cdot (S + 0.02) \]

### Step 4: Substitute the value of Ksp
We know that \( K_{sp} \) for BaSO₄ is \( 4 \times 10^{-10} \):
\[ 4 \times 10^{-10} = S \cdot (S + 0.02) \]

### Step 5: Expand and rearrange the equation
Expanding the equation gives:
\[ 4 \times 10^{-10} = S^2 + 0.02S \]
Rearranging this gives us a quadratic equation:
\[ S^2 + 0.02S - 4 \times 10^{-10} = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 0.02 \), and \( c = -4 \times 10^{-10} \).

Calculating the discriminant:
\[ b^2 - 4ac = (0.02)^2 - 4(1)(-4 \times 10^{-10}) \]
\[ = 0.0004 + 1.6 \times 10^{-9} \]
\[ = 0.0004 + 0.0000016 = 0.0004016 \]

Now substituting back into the quadratic formula:
\[ S = \frac{-0.02 \pm \sqrt{0.0004016}}{2} \]
Calculating the square root:
\[ \sqrt{0.0004016} \approx 0.02004 \]
Now substituting:
\[ S = \frac{-0.02 \pm 0.02004}{2} \]

Calculating the two possible values:
1. \( S = \frac{0.00004}{2} = 0.00002 \) (valid solution)
2. \( S = \frac{-0.04004}{2} \) (discard as it is negative)

Thus, the solubility \( S \) is approximately:
\[ S \approx 2 \times 10^{-5} \, \text{mol/L} \]

### Final Answer:
The solubility of BaSO₄ in the presence of 0.02 N H₂SO₄ is approximately \( 2 \times 10^{-5} \, \text{mol/L} \).

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