chemistry

how do i calculate the molar solubility of AgBr in .10M NaBr?

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  1. AgBr<==> Ag^+ + Br^-
    Ksp = (Ag^+)(Br^-) = ??look up in your text or tables.
    solubility AgBr = y
    (Ag^+) = y
    (Br^-) = y from AgBr and 0.10 from NaBr so (Br^-) = y+0.10
    Set up in Ksp expression above and solve for y.

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