Solubility products of `ACO_(3), BSO_(4)`, and `ASO_(4)`, are `4 xx 10^(-10), 6 xx 10^(-10)` and `8 xx 10^(-10)` respectively. The solubility product of `BCO_(3)`, is x x `10^(-10)`, What is x ?

Answer

To solve the problem, we need to find the value of \( x \) in the solubility product of \( BCO_3 \) given the solubility products of \( ACO_3 \), \( BSO_4 \), and \( ASO_4 \).

### Step 1: Write the dissociation equations
1. For \( ACO_3 \):
\[ ACO_3 \rightleftharpoons A^{2+} + CO_3^{2-} \]
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [A^{2+}][CO_3^{2-}] \]

2. For \( BSO_4 \):
\[ BSO_4 \rightleftharpoons B^{2+} + SO_4^{2-} \]
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [B^{2+}][SO_4^{2-}] \]

3. For \( ASO_4 \):
\[ ASO_4 \rightleftharpoons A^{2+} + SO_4^{2-} \]
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [A^{2+}][SO_4^{2-}] \]

### Step 2: Use the given solubility products
- \( K_{sp}(ACO_3) = 4 \times 10^{-10} \)
- \( K_{sp}(BSO_4) = 6 \times 10^{-10} \)
- \( K_{sp}(ASO_4) = 8 \times 10^{-10} \)

### Step 3: Relate the solubility products
From the dissociation of \( ACO_3 \) and \( BSO_4 \), we can set up a relationship between the solubility products:
\[ \frac{[CO_3^{2-}]}{[SO_4^{2-}]} = \frac{K_{sp}(ACO_3)}{K_{sp}(ASO_4)} \]
Substituting the values:
\[ \frac{[CO_3^{2-}]}{[SO_4^{2-}]} = \frac{4 \times 10^{-10}}{8 \times 10^{-10}} = \frac{1}{2} \]
This implies:
\[ [SO_4^{2-}] = 2[CO_3^{2-}] \]

### Step 4: Write the solubility product for \( BCO_3 \)
For \( BCO_3 \):
\[ BCO_3 \rightleftharpoons B^{2+} + CO_3^{2-} \]
Thus, the solubility product \( K_{sp}(BCO_3) \) is:
\[ K_{sp}(BCO_3) = [B^{2+}][CO_3^{2-}] \]

### Step 5: Relate \( [B^{2+}] \) to \( [SO_4^{2-}] \)
From \( BSO_4 \):
\[ K_{sp}(BSO_4) = [B^{2+}][SO_4^{2-}] \]
Substituting for \( [SO_4^{2-}] \):
\[ K_{sp}(BSO_4) = [B^{2+}](2[CO_3^{2-}]) \]
Thus:
\[ 6 \times 10^{-10} = [B^{2+}](2[CO_3^{2-}]) \]
This gives:
\[ [B^{2+}] = \frac{6 \times 10^{-10}}{2[CO_3^{2-}]} = \frac{3 \times 10^{-10}}{[CO_3^{2-}]} \]

### Step 6: Substitute back to find \( K_{sp}(BCO_3) \)
Substituting \( [B^{2+}] \) into the \( K_{sp}(BCO_3) \):
\[ K_{sp}(BCO_3) = \left(\frac{3 \times 10^{-10}}{[CO_3^{2-}]}\right)[CO_3^{2-}] = 3 \times 10^{-10} \]

### Conclusion
Thus, the solubility product of \( BCO_3 \) is:
\[ K_{sp}(BCO_3) = 3 \times 10^{-10} \]
Therefore, \( x = 3 \).