MHT CET · Chemistry · Ionic Equilibrium
The solubility of \(\mathrm{AgCl}\) in it's solution is \(1.25 \times 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3}\). What is solubility product of \(\mathrm{AgCl}\) ?
- A \(1.56 \times 10^{-10}\)
- B \(3.50 \times 10^{-6}\)
- C \(1.10 \times 10^{-5}\)
- D \(2.53 \times 10^{-3}\)
Answer & Solution
Correct Answer
(A) \(1.56 \times 10^{-10}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{AgCl}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)
\(\therefore \mathrm{Ksp}=\mathrm{S}^2=\left(1.25 \times 10^{-5}\right)^2=1.56 \times 10^{-10}\)
\(\therefore \mathrm{Ksp}=\mathrm{S}^2=\left(1.25 \times 10^{-5}\right)^2=1.56 \times 10^{-10}\)
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