AP EAMCET · Chemistry · Ionic Equilibrium
At \(25^{\circ} \mathrm{C}\), the solubility product of \(\mathrm{MCl}\) is \(1 \times 10^{-10}\). What is its molar solubility in \(0.1 \mathrm{M}\) \(\mathrm{NaCl}\) solution at same temperature?
- A \(0.1\)
- B \(0.05\)
- C \(10^{-9}\)
- D \(10^{-5}\)
Answer & Solution
Correct Answer
(C) \(10^{-9}\)
Step-by-step Solution
Detailed explanation
\[ K_{\text {sp }} \text { of } M C l=1 \times 10^{-10} \text {. } \] Let the molar solubility of \(\mathrm{MCl}\) in \(0.1 \mathrm{M} \mathrm{NaCl}\) be ' \(S\) ' \(\mathrm{mol} \mathrm{L}^{-1}\). \[ \mathrm{MCl} \longrightarrow \mathrm{M}^{+}+\mathrm{Cl}^{-} \] The…
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