MHT CET · Chemistry · Ionic Equilibrium
The solubility of AgBr is \(7.1 \times 10^{-7} \mathrm{~mol~} \mathrm{dm}^{-3}\). Calculate it's solubility product at the same temperature.
- A \(7.08 \times 10^{-13}\)
- B \(3.67 \times 10^{-13}\)
- C \(5.89 \times 10^{-13}\)
- D \(5.04 \times 10^{-13}\)
Answer & Solution
Correct Answer
(D) \(5.04 \times 10^{-13}\)
Step-by-step Solution
Detailed explanation
\(K_{sp} = s^2\) \(K_{sp} = (7.1 \times 10^{-7})^2 = 5.041 \times 10^{-13}\)
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