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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 \(\mathrm{AgCl}\) ?

  1. A \(1.56 \times 10^{-10}\)
  2. B \(3.50 \times 10^{-6}\)
  3. C \(1.10 \times 10^{-5}\)
  4. D \(2.53 \times 10^{-3}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1.56 \times 10^{-10}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{AgCl}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-}\)
\(\therefore \mathrm{Ksp}=\mathrm{S}^2=\left(1.25 \times 10^{-5}\right)^2=1.56 \times 10^{-10}\)