The solubility product of a sparingly soluble salt $BA$ is $4 \times 10^{-13}$. Calculate the $[A^-]$ if $[B^+]$ is $1 \times 10^{-6} \text{ M}$.

  • A
    $2 \times 10^{-7} \text{ M}$
  • B
    $3 \times 10^{-7} \text{ M}$
  • C
    $4 \times 10^{-7} \text{ M}$
  • D
    $5 \times 10^{-7} \text{ M}$

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