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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|---|---|
| $AgBrO_3$ | $2.5 \times 10^{-10}$ |
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| $CaCO_3$ | $2.7 \times 10^{-18}$ |
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