If the solubility of `CaSO_(4)` in `H_(2)O` is `10^(-5)M`, Calculate the solubility in `0.005M` solution of `H_(2)SO_(4)`.

Explanation

To solve the problem of calculating the solubility of \( \text{CaSO}_4 \) in a \( 0.005 \, M \) solution of \( \text{H}_2\text{SO}_4 \), we will follow these steps:

### Step 1: Calculate the \( K_{sp} \) of \( \text{CaSO}_4 \)

Given that the solubility of \( \text{CaSO}_4 \) in water is \( 10^{-5} \, M \), we can write the dissociation equation:

\[ \text{CaSO}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

From the dissociation, we see that:

- The concentration of \( \text{Ca}^{2+} \) ions = \( S = 10^{-5} \, M \)
- The concentration of \( \text{SO}_4^{2-} \) ions = \( S = 10^{-5} \, M \)

Thus, the solubility product \( K_{sp} \) is given by:

\[ K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] = (10^{-5})(10^{-5}) = 10^{-10} \]

### Step 2: Consider the dissociation of \( \text{H}_2\text{SO}_4 \)

The dissociation of \( \text{H}_2\text{SO}_4 \) can be represented as:

\[ \text{H}_2\text{SO}_4 \rightarrow 2 \text{H}^+ + \text{SO}_4^{2-} \]

In a \( 0.005 \, M \) solution of \( \text{H}_2\text{SO}_4 \):

- Initial concentration of \( \text{H}_2\text{SO}_4 \) = \( 0.005 \, M \)
- After complete dissociation, the concentration of \( \text{SO}_4^{2-} \) ions = \( 0.005 \, M \)

### Step 3: Apply the common ion effect

The presence of \( \text{SO}_4^{2-} \) ions from \( \text{H}_2\text{SO}_4 \) will suppress the solubility of \( \text{CaSO}_4 \) due to the common ion effect. We can denote the new solubility of \( \text{CaSO}_4 \) in the presence of the common ion as \( S' \).

Using the formula for solubility in the presence of a common ion:

\[ S' = \frac{K_{sp}}{C^n} \]

Where:
- \( C \) = concentration of the common ion \( \text{SO}_4^{2-} = 0.005 \, M \)
- \( n \) = number of \( \text{SO}_4^{2-} \) ions = 1

Substituting the values:

\[ S' = \frac{10^{-10}}{(0.005)^1} = \frac{10^{-10}}{0.005} = 2 \times 10^{-8} \, M \]

### Final Answer

The solubility of \( \text{CaSO}_4 \) in \( 0.005 \, M \) solution of \( \text{H}_2\text{SO}_4 \) is \( 2 \times 10^{-8} \, M \).

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