The solubility product of `BaSO_(4) and BaCrO_(4)` at `25^(@)C` are `1xx10^(-10)` respectively. Calculate the simultaneous solubilities of `BaSO_(4) and BaCrO_(4).`

Answer

To solve the problem of calculating the simultaneous solubilities of BaSO₄ and BaCrO₄, we can follow these steps:

### Step 1: Write the dissociation equations and expressions for solubility products.

For Barium Sulfate (BaSO₄):
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]
The solubility product expression is:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]

For Barium Chromate (BaCrO₄):
\[ \text{BaCrO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{CrO}_4^{2-} (aq) \]
The solubility product expression is:
\[ K_{sp} = [\text{Ba}^{2+}][\text{CrO}_4^{2-}] \]

### Step 2: Define the solubility of each salt.

Let the solubility of BaSO₄ be \( S_1 \) and the solubility of BaCrO₄ be \( S_2 \). Since both salts produce Ba²⁺ ions, we can express their concentrations as:
\[ [\text{Ba}^{2+}] = S_1 + S_2 \]

### Step 3: Substitute into the Ksp expressions.

For BaSO₄:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = (S_1 + S_2)S_1 \]
Given \( K_{sp} = 1 \times 10^{-10} \):
\[ 1 \times 10^{-10} = (S_1 + S_2)S_1 \quad \text{(1)} \]

For BaCrO₄:
\[ K_{sp} = [\text{Ba}^{2+}][\text{CrO}_4^{2-}] = (S_1 + S_2)S_2 \]
Given \( K_{sp} = 1 \times 10^{-10} \):
\[ 1 \times 10^{-10} = (S_1 + S_2)S_2 \quad \text{(2)} \]

### Step 4: Solve the equations simultaneously.

From equations (1) and (2), we can set them equal to each other:
\[ (S_1 + S_2)S_1 = (S_1 + S_2)S_2 \]
Assuming \( S_1 + S_2 \neq 0 \), we can divide both sides by \( S_1 + S_2 \):
\[ S_1 = S_2 \]
Let \( S_1 = S_2 = S \). Then substituting back into either equation gives:
\[ 1 \times 10^{-10} = (2S)S = 2S^2 \]
Solving for \( S \):
\[ S^2 = \frac{1 \times 10^{-10}}{2} = 5 \times 10^{-11} \]
\[ S = \sqrt{5 \times 10^{-11}} = \sqrt{5} \times 10^{-5.5} \approx 7.07 \times 10^{-6} \text{ mol/L} \]

### Final Answer:
The simultaneous solubility of BaSO₄ and BaCrO₄ is approximately \( 7.07 \times 10^{-6} \text{ mol/L} \).

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