Sigh, I am starting to think that every answer on this answer key is wrong... If someone can confirm that I did this question correctly, I will just disregard the answer key for the rest of the questions.

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Calculate the Ag^{+} concentration required to start a precipitation of AgBrO_{3} from a solution containing 0.50g of NaBrO_{3} per 0.300 litre. Ksp of AgBrO_{3} = 5.3(10^{-5})

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[BrO_{3}^{-}] = [NaBrO_{3}] = 0.50g(1mol/127.9g)(1/0.300L) = 0.013M

Ksp = [Ag^{+}][BrO_{3}^{-}]

5.3(10^{-5}) = [Ag^{+}](0.013M)

4.1(10^{-3})M = [Ag^{+}]

Answer key says it's 4.5(10^{-11})M...