[dagr8est]

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 AgBrO₃ from a solution containing 0.50g of NaBrO₃ per 0.300 litre.  Ksp of AgBrO₃ = 5.3(10^-5)

***

[BrO₃^-] = [NaBrO₃] = 0.50g(1mol/127.9g)(1/0.300L) = 0.013M

Ksp = [Ag⁺][BrO₃^-]
5.3(10^-5) = [Ag⁺](0.013M)
4.1(10^-3)M = [Ag⁺]

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

[Borek]

Seems to me it should be 4.8*10^-3, so both answers are wrong. Check molar mass of sodium bromate.

[dagr8est]

Oh yeah whoops, I forgot to include sodium.  I agree it should be 4.8(10^-3)M then.  I guess this answer key is just completely wrong then.  Thanks for the help on both questions.

[AWK]

MM  of NaBrO3 is ~ 151