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Problems · Problem 6.27

Q.The values of Ksp of two sparingly soluble salts Ni(OH)2 and AgCN are 2.0 × 10⁻¹⁵ and 6 × 10⁻¹⁷ respectively. Which salt is more soluble? Explain.

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Compare molar solubilities by writing dissociation equations and expressing KspK_{sp} in terms of solubility ss; stoichiometry matters. Ni(OH)₂ is more soluble (s=7.94×10−6s = 7.94 \times 10^{-6} M) than AgCN (s=7.75×10−9s = 7.75 \times 10^{-9} M), even though its KspK_{sp} is smaller.

Note

The NCERT textbook's printed solution shows S2=0.58×10−4S_2 = 0.58 \times 10^{-4} for Ni(OH)2\text{Ni(OH)}_2 — a calculation slip: from 4S23=2×10−154S_2^3 = 2 \times 10^{-15}, S2=(5.0×10−16)1/3=7.94×10−6S_2 = (5.0 \times 10^{-16})^{1/3} = 7.94 \times 10^{-6} M, as computed here. The book's printed verdict — Ni(OH)₂ is more soluble than AgCN — still holds either way.

The solubility product constant KspK_{sp} tells us about equilibrium between a sparingly soluble salt and its ions in solution. A common mistake is to assume that a larger KspK_{sp} always means greater solubility — but that's only true when comparing salts with identical stoichiometry. The relationship between KspK_{sp} and molar solubility depends on how many ions form when one formula unit dissolves.

When Ni(OH)₂ dissolves, it produces one Ni²⁺ ion and two OH⁻ ions. When AgCN dissolves, it produces one Ag⁺ and one CN⁻. These different stoichiometries mean we must calculate the actual molar solubility ss for each salt before comparing.

Step-by-step comparison

1. Write the dissociation equilibrium for Ni(OH)₂

Ni(OH)2(s)⇌Ni2+(aq)+2OH−(aq)\text{Ni(OH)}_2(s) \rightleftharpoons \text{Ni}^{2+}(aq) + 2\text{OH}^-(aq)

If the molar solubility is ss mol/L, then at equilibrium:

  • [Ni2+]=s[\text{Ni}^{2+}] = s
  • [OH−]=2s[\text{OH}^-] = 2s (because two hydroxide ions form per formula unit)

2. Express KspK_{sp} for Ni(OH)₂ in terms of ss

Ksp=[Ni2+][OH−]2=s⋅(2s)2=4s3K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2 = s \cdot (2s)^2 = 4s^3

Given Ksp=2.0×10−15K_{sp} = 2.0 \times 10^{-15}:

4s3=2.0×10−154s^3 = 2.0 \times 10^{-15}

s3=5.0×10−16s^3 = 5.0 \times 10^{-16}

s=(5.0×10−16)1/3=7.94×10−6 Ms = (5.0 \times 10^{-16})^{1/3} = 7.94 \times 10^{-6} \text{ M}

3. Write the dissociation equilibrium for AgCN

AgCN(s)⇌Ag+(aq)+CN−(aq)\text{AgCN}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{CN}^-(aq)

If the molar solubility is ss mol/L:

  • [Ag+]=s[\text{Ag}^+] = s
  • [CN−]=s[\text{CN}^-] = s

4. Express KspK_{sp} for AgCN in terms of ss

Ksp=[Ag+][CN−]=s⋅s=s2K_{sp} = [\text{Ag}^+][\text{CN}^-] = s \cdot s = s^2 …

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