ExamBro
ExamBro
MHT CET · Chemistry · Ionic Equilibrium

Solubility product of \(\mathrm{AgBr}\) is \(4.9 \times 10^{-13}\). What is its solubility?

  1. A \(2.4 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)
  2. B \(3.2 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)
  3. C \(4.9 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)
  4. D \(7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \mathrm{K}_{\mathrm{sp}}(\mathrm{AgBr})=\mathrm{S}^2 \\ & 4.9 \times 10^{-13}=\mathrm{S}^2 \\ & \mathrm{~S}=\sqrt{4.9 \times 10^{-13}} \\ & \mathrm{~S}=7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}\)