# Chemistry

Suppose 26.0g of sodium bromide is dissolved in 150mL of a 0.60M aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the sodium bromide is dissolved in it.

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1. NaBr + AgNO3 ==> AgBr(s) + NaNO3
mols NaBr = grams/molar mass = estimated 0.25
mols AgNO3 = M x L = estd 0.09
So all of the AgNO3 will be used and you will have 0.25-0.09 = 0.163 mols NaBr unreacted.
(NaBr) = (Br^-) = mols/L = 0.163/0.150 =?
I have assumed that the amount of Br^- furnished by the slightly soluble AgBr is negligible. I'm sure that is so although I didn't calculate how much that would be.

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