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NCERT Exemplar · Q12

Q.In which of the following solvents is silver chloride most soluble?

(i) 0.1 mol dm^-3 AgNO3 solution
(ii) 0.1 mol dm^-3 HCl solution
(iii) H2O
(iv) Aqueous ammonia
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Silver chloride is most soluble in aqueous ammonia because ammonia forms a stable complex ion with silver ions, effectively removing them from solution and shifting the dissolution equilibrium. The correct option is (iv).

The solubility of an ionic compound like silver chloride (AgClAgCl) in a solvent depends on the extent to which its dissolution equilibrium can be shifted. Silver chloride is a sparingly soluble salt, meaning only a small amount dissolves in water. Its dissolution can be represented by the equilibrium:

AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)

The solubility product constant (KspK_{sp}) for this equilibrium is given by Ksp=[Ag+][Cl−]K_{sp} = [Ag^+][Cl^-]. A higher concentration of Ag+Ag^+ or Cl−Cl^- ions in the solution (due to other sources) will shift this equilibrium to the left, decreasing the solubility of AgClAgCl. Conversely, if Ag+Ag^+ or Cl−Cl^- ions are removed from the solution, the equilibrium will shift to the right, increasing the solubility.

Let's analyze each solvent option:

  1. Solubility in H2_2O (option (iii))

    In pure water, AgClAgCl dissolves to a small extent, establishing the equilibrium:

    AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)

    If ss is the molar solubility of AgClAgCl in water, then [Ag+]=s[Ag^+] = s and [Cl−]=s[Cl^-] = s.

    So, Ksp=s2K_{sp} = s^2, and s=Ksps = \sqrt{K_{sp}}. This serves as our baseline for comparison. For AgClAgCl, KspK_{sp} is approximately 1.8×10−101.8 \times 10^{-10} at 25∘C25^\circ C, so s≈1.34×10−5s \approx 1.34 \times 10^{-5} mol dm−3^{-3}.

  2. Solubility in 0.1 mol dm−3^{-3} AgNO3_3 solution (option (i))

    Silver nitrate (AgNO3AgNO_3) is a strong electrolyte and dissociates completely in water:

    AgNO3(aq)→Ag+(aq)+NO3−(aq)AgNO_3(aq) \rightarrow Ag^+(aq) + NO_3^-(aq)

    This solution introduces a significant concentration of Ag+Ag^+ ions (0.10.1 mol dm−3^{-3}) into the solvent. According to Le Chatelier's principle, the presence of a common ion (Ag+Ag^+) will shift the dissolution equilibrium of AgClAgCl to the left:

    AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)

    This reduces the solubility of AgClAgCl compared to its solubility in pure water. If s′s' is the solubility in AgNO3AgNO_3, then [Ag+]≈0.1[Ag^+] \approx 0.1 M (since s′s' will be very small) and [Cl−]=s′[Cl^-] = s'.

    Ksp=[Ag+][Cl−]≈(0.1)(s′)K_{sp} = [Ag^+][Cl^-] \approx (0.1)(s').

    s′=Ksp/0.1s' = K_{sp}/0.1. This value will be much smaller than Ksp\sqrt{K_{sp}}.

  3. Solubility in 0.1 mol dm−3^{-3} HCl solution (option (ii))

    Hydrochloric acid (HClHCl) is a strong acid and dissociates completely in water:

    HCl(aq)→H+(aq)+Cl−(aq)HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)

    This solution introduces a significant concentration of Cl−Cl^- ions (0.10.1 mol dm−3^{-3}) into the solvent. Similar to the AgNO3AgNO_3 case, the presence of a common ion (Cl−Cl^-) will shift the dissolution equilibrium of AgClAgCl to the left:

    AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)

    This also reduces the solubility of AgClAgCl compared to its solubility in pure water. If s′′s'' is the solubility in HClHCl, then [Ag+]=s′′[Ag^+] = s'' and [Cl−]≈0.1[Cl^-] \approx 0.1 M.

    Ksp=[Ag+][Cl−]≈(s′′)(0.1)K_{sp} = [Ag^+][Cl^-] \approx (s'')(0.1).

    s′′=Ksp/0.1s'' = K_{sp}/0.1. This value will also be much smaller than Ksp\sqrt{K_{sp}}. …

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