MHT CET · Chemistry · Ionic Equilibrium
Solubility product of \(\mathrm{AgBr}\) is \(4.9 \times 10^{-13}\). What is its solubility?
- A \(2.4 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)
- B \(3.2 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)
- C \(4.9 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)
- D \(7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)
Answer & Solution
Correct Answer
(D) \(7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{K}_{\mathrm{sp}}(\mathrm{AgBr})=\mathrm{S}^2 \\ & 4.9 \times 10^{-13}=\mathrm{S}^2 \\ & \mathrm{~S}=\sqrt{4.9 \times 10^{-13}} \\ & \mathrm{~S}=7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}\)
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