If the solubility of $PbBr_2$ is $S \, mol/L$,and it undergoes $100\%$ ionization,then the solubility product constant $(K_{sp})$ is equal to:

  • A
    $2S^3$
  • B
    $4S^2$
  • C
    $4S^3$
  • D
    $2S^4$

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