The solubility of $AgBr$ with solubility product $5.0 \times 10^{-13}$ at $298 \ K$ in $0.1 \ M$ $NaBr$ solution would be:

  • A
    $7 \times 10^{-6} \ M$
  • B
    $5 \times 10^{-12} \ M$
  • C
    $5 \times 10^{-14} \ M$
  • D
    $5 \times 10^{-6} \ M$

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