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AgBr + Ag(NO)3 ----------> Ag+ + Br-

Disregard Ag(NO)3 b/c of the common ion Ag and the fact that (NO)3 is a spectator ion.

You use the Ksp of AgBr from the index in your book that equals 5.0*10^-13

If you look at your above equation, Ksp = [Ag] [Br]

You know Ksp = 5.0*10^-13

You have also been given the concentration of Ag from Ag(NO)3 = 3.0*10^-2. You use this number because remember NO3 is always a spectator ion so [Ag] will equal the [Ag(NO)3]

You equation should look like this. (Forget about doing the I.C.E on this one. Its not necessary.

5.0*10^-13-= [3.0*10^-2] [Br]

Simply solve for [Br] by dividing 5.0*10^-13 by 3.0*10^-2.

[Br] = 1.66*10^-11. Round for sig figs to = 1.7*10-11

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13y ago

Ksp of AgBr is 7.7 × 10−13

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Q: Calculate the molar solubility of AgBr in a 3.0 x 10-2 M silver nitrate AgNO3 solution?
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