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AIIMS 2002
Exam
AIIMS
Year
2002
Subject
Chemistry
Chapter
Ionic Equilibrium

AIIMS 2002 · Chemistry · Ionic Equilibrium

The solubility of CuBr is 2 × 10^ -4 ~mol / L at 25^ C . The K_ s p...

The solubility of CuBr\mathrm{CuBr} is 2×10−4 mol/L2 \times 10^{-4} \mathrm{~mol} / \mathrm{L} at 25∘C25^{\circ} \mathrm{C}. The KspK_{s p} value for CuBr\mathrm{CuBr} is

Options

  1. A4×10−8 mol2 L−24 \times 10^{-8} \mathrm{~mol}^2 \mathrm{~L}^{-2}
  2. B4×10−4 mol2 L−24 \times 10^{-4} \mathrm{~mol}^2 \mathrm{~L}^{-2}
  3. C4×10−11 mol2 L−24 \times 10^{-11} \mathrm{~mol}^2 \mathrm{~L}^{-2}
  4. D4×10−15 mol2 L−24 \times 10^{-15} \mathrm{~mol}^2 \mathrm{~L}^{-2}

Correct answer

A. 4×10−8 mol2 L−24 \times 10^{-8} \mathrm{~mol}^2 \mathrm{~L}^{-2}

Step-by-step solution

CuBr→Cu++Br−\mathrm{CuBr} \rightarrow \mathrm{Cu}^{+}+\mathrm{Br}^{-} Solubility of CuBr\mathrm{CuBr} is 2×10−4 mol L−12 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} Therefore, solubility of Cu+=2×10−4 mol L−1\mathrm{Cu}^{+}=2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} ⇒\Rightarrow Solubility product =2×10−4×2×10−4=2 \times 10^{-4} \times 2 \times 10^{-4} =4×10−8 mol2 L−2=4 \times 10^{-8} \mathrm{~mol}^2 \mathrm{~L}^{-2}
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