WBJEE · Chemistry · Ionic Equilibrium
Increasing order of solubility of AgCl in (i) \(\mathrm{H}_2 \mathrm{O}\), (ii) 1 M NaCl (aq.), (iii) \(1 \mathrm{M}\) \(\mathrm{CaCl}_2\) (aq.) and (iv) \(1 \mathrm{M}\) \(\mathrm{NaNO}_3\) (aq.) solution
- A \(\mathrm{CaCl}_2 \lt \mathrm{NaNO}_3 \lt \mathrm{NaCl} \lt \mathrm{H}_2 \mathrm{O}\)
- B \(\mathrm{CaCl}_2\gt\mathrm{H}_2 \mathrm{O}\gt\mathrm{NaCl}\gt\mathrm{NaNO}_3\)
- C \(\mathrm{CaCl}_2\gt\mathrm{NaCl}\gt\mathrm{H}_2 \mathrm{O}\gt\mathrm{NaNO}_3\)
- D \(\mathrm{CaCl}_2 \lt \mathrm{NaCl} \lt \mathrm{H}_2 \mathrm{O} \lt \mathrm{NaNO}_3\)
Answer & Solution
Correct Answer
(D) \(\mathrm{CaCl}_2 \lt \mathrm{NaCl} \lt \mathrm{H}_2 \mathrm{O} \lt \mathrm{NaNO}_3\)
Step-by-step Solution
Detailed explanation
\(\text { In } \mathrm{H}_2 \mathrm{O} \text { solubility }=\sqrt{\mathrm{K}_{\mathrm{sp}}}\) In 1 M NaCl and in \(1 \mathrm{M} \mathrm{CaCl}_2\) more the [ \(\mathrm{Cl}^{-}\)] less is the solubility Hence…
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