Q53 / 180 · AIIMS 2002 · Chemistry

The solubility of $\mathrm{CuBr}$ is $2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$ at $25^{\circ} \mathrm{C}$. The $K_{s p}$ value for $\mathrm{CuBr}$ is

  • Correct answer
    A
    $4 \times 10^{-8} \mathrm{~mol}^2 \mathrm{~L}^{-2}$
  • B
    $4 \times 10^{-4} \mathrm{~mol}^2 \mathrm{~L}^{-2}$
  • C
    $4 \times 10^{-11} \mathrm{~mol}^2 \mathrm{~L}^{-2}$
  • D
    $4 \times 10^{-15} \mathrm{~mol}^2 \mathrm{~L}^{-2}$

Solution

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

Hence, the correct answer is option: 'A'.

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