Question

The values of \( a \) and \( b \) in the solubility product equation for barium phosphate are:

• A. \( 7, 5 \)
• B. \( 5, 7 \)
• C. \( 5, 5 \)
• D. \( 7, 7 \)

Hint:

For a compound \( A_mB_n \), the solubility product follows \( K_{sp} = [A]^{m} [B]^{n} \).

Answer 1

The Correct Option is B

Step 1: Dissociation of Barium Phosphate The chemical formula of barium phosphate is: \[ Ba_3(PO_4)_2 \] It dissociates as: \[ Ba_3(PO_4)_2 \rightleftharpoons 3Ba^{2+} + 2PO_4^{3-} \] Step 2: Express Ion Concentrations in Terms of \( x \)
- The solubility is \( x \) g per 100 mL.
- The molar solubility is \( \frac{x}{M} \).
From the dissociation:
- \( [Ba^{2+}] = 3 \times \frac{x}{M} \)
- \( [PO_4^{3-}] = 2 \times \frac{x}{M} \) Step 3: Write the Solubility Product Expression \[ K_{sp} = [Ba^{2+}]^3 \times [PO_4^{3-}]^2 \] \[ K_{sp} = \left( 3 \times \frac{x}{M} \right)^3 \times \left( 2 \times \frac{x}{M} \right)^2 \] \[ = 27 \times \left( \frac{x}{M} \right)^3 \times 4 \times \left( \frac{x}{M} \right)^2 \] \[ = 108 \times \left( \frac{x}{M} \right)^5 \] \[ = 1.08 \times 10^2 \times \left( \frac{x}{M} \right)^5 \] Step 4: Compare with the Given Equation \[ K_{sp} = 1.08 \times \left( \frac{x}{M} \right)^a \times (10)^b \] Thus, we get: \[ a = 5, \quad b = 7 \]