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Problems · Problem 6.28

Q.Calculate the molar solubility of Ni(OH)2 in 0.10 M NaOH. The ionic product of Ni(OH)2 is 2.0 × 10⁻¹⁵.

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In a basic solution, the common ion OH⁻ from NaOH suppresses the dissolution of Ni(OH)₂. Using the solubility product Ksp=2.0×10−15K_{sp} = 2.0 \times 10^{-15} and the fixed [OH−]=0.10 M[OH^-] = 0.10\ \text{M}, the molar solubility is 2.0×10−13 M2.0 \times 10^{-13}\ \text{M}.

The key here is the common ion effect. Ni(OH)₂ is a sparingly soluble salt. In pure water, its dissolution is limited only by its KspK_{sp}. But when we add NaOH, a strong base that fully dissociates, we flood the solution with OH⁻ ions. According to Le Chatelier’s principle, the equilibrium

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

shifts to the left — solubility drops dramatically. The KspK_{sp} expression is the same, but now the OH⁻ concentration is no longer coming solely from the salt; it’s dominated by the added base.

Ksp=[Ni2+][OH−]2K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2

Let’s work through the calculation.

  1. Set up the equilibrium. Let ss be the molar solubility of Ni(OH)₂ in mol/L. That means ss moles of Ni(OH)₂ dissolve per litre, producing ss mol/L of Ni²⁺ and 2s2s mol/L of OH⁻ from the salt itself. But we already have 0.10 M OH⁻ from NaOH. So the total hydroxide concentration is:

[OH−]=0.10+2s[\text{OH}^-] = 0.10 + 2s

Since ss will be extremely small (we’ll see why in a moment), 2s≪0.102s \ll 0.10, so we can safely approximate:

[OH−]≈0.10 M[\text{OH}^-] \approx 0.10\ \text{M}

  1. Write the KspK_{sp} expression and substitute. The ionic product given is 2.0×10−152.0 \times 10^{-15}, which is the KspK_{sp}:

Ksp=[Ni2+][OH−]2=(s)(0.10)2K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2 = (s)(0.10)^2

So:

2.0×10−15=s×(0.01)2.0 \times 10^{-15} = s \times (0.01)

  1. Solve for ss: s=2.0×10−150.01=2.0×10−13 Ms = \frac{2.0 \times 10^{-15}}{0.01} = 2.0 \times 10^{-13}\ \text{M} …

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