Question

Solubility product of the sparingly soluble salt \(AgBrO_3\) in aqueous medium is \(9.3 \times 10^{-10}\). Calculate the mass in gram of \(AgBrO_3\) present in \(100 \, ml\) of its saturated solution. (Molar mass of \(AgBrO_3 = 236 \, g \, mol^{-1}\))

• A. \(3.0495 \times 10^{-4}\)
• B. \(4.962 \times 10^{-4}\)
• C. \(6.248 \times 10^{-5}\)
• D. \(7.198 \times 10^{-4}\)

Hint:

For salts of type \(AB\), \(K_{sp} = s^2\). First find solubility in mol/L, then convert into mass using molar mass.

Answer 1

The Correct Option is D

Step 1: Write the dissociation equilibrium.
The salt \(AgBrO_3\) dissociates in water as:
\[ AgBrO_3 (s) \rightleftharpoons Ag^+ + BrO_3^- \] Let the solubility of the salt be \(s\).
Then:
\[ [Ag^+] = s, \quad [BrO_3^-] = s \]

Step 2: Write the expression for solubility product.
\[ K_{sp} = [Ag^+][BrO_3^-] \] \[ K_{sp} = s \times s = s^2 \]

Step 3: Calculate solubility \(s\).
Given:
\[ K_{sp} = 9.3 \times 10^{-10} \] \[ s^2 = 9.3 \times 10^{-10} \] \[ s = \sqrt{9.3 \times 10^{-10}} \] \[ s = 3.05 \times 10^{-5} \, mol \, L^{-1} \]

Step 4: Calculate moles in \(100 \, ml\) solution.
\[ 100 \, ml = 0.1 \, L \] \[ \text{Moles} = s \times V \] \[ \text{Moles} = 3.05 \times 10^{-5} \times 0.1 \] \[ \text{Moles} = 3.05 \times 10^{-6} \, mol \]

Step 5: Calculate mass of dissolved salt.
\[ \text{Mass} = \text{Moles} \times \text{Molar mass} \] \[ \text{Mass} = 3.05 \times 10^{-6} \times 236 \] \[ \text{Mass} = 7.198 \times 10^{-4} \, g \]

Step 6: Select the correct option.
The calculated value matches option (D).
Final Answer:
The mass of \(AgBrO_3\) in \(100 \, ml\) saturated solution is:
\[ \boxed{7.198 \times 10^{-4} \, g} \]