The solubility in water of a sparingly soluble salt $AB_2$ is $1.0 \times 10^{-5} \ mol \ L^{-1}$. Its solubility product constant will be

  • A
    $4 \times 10^{-15}$
  • B
    $4 \times 10^{-10}$
  • C
    $1 \times 10^{-15}$
  • D
    $1 \times 10^{-10}$

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The solubility product of a sparingly soluble salt $BA$ is $6.4 \times 10^{-13}$. Calculate its solubility in $\text{g dm}^{-3}$. The molar mass of the salt is $190 \text{ g mol}^{-1}$.

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