Problem: A solution contains 0.018 mole each of I -, Br -, and Cl -. When the solution is mixed with 200. mL of 0.24 M AgNO3, what mass of AgCl(s) precipitates out, and what is [Ag +]? Assume no volume change.AgI: Ksp = 1.5 X 10 -16AgBr: Ksp = 5.0 X 10 -13AgCl: Ksp = 1.6 X 10 -10

FREE Expert Solution

The lower ksp compound will precipitate first: 

AgI: Ksp = 1.5 X 10 -16
AgBr: Ksp = 5.0 X 10 -13
AgCl: Ksp = 1.6 X 10 -10 

AgI > AgBr > AgCl


AgCl will be the last to precipitate. 

1 Ag+ : 1 I- : 1 Br- : 1 Cl- 

This means the amount of each halide can be used to calculate how much Ag+ will be left before Cl- could react. 

Mol Ag+= 0.24 mol1 L×200 mL×10-3 L1 mL

Mol Ag+= 0.048 mol 


Calculate mole Ag+ that Cl- could react with: 

mol Ag+ left = mol Ag+initial - mol I- - mol Br-mol Ag+ left = 0.048 mol - 0.018 mol - 0.018 mol

mol Ag+ left = 0.012 mol


We still have 0.018 mol Cl-, so Ag+ will be limiting since it is less.

This also means: 

 Ag+ Cl- →AgCl

0.012 mol Ag+× 1 mol AgCl1  mol Ag+

= 0.012 mol AgCl


Since Ag+ is limiting, there would be no more Ag+ left. 


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Problem Details

A solution contains 0.018 mole each of I -, Br -, and Cl -. When the solution is mixed with 200. mL of 0.24 M AgNO3, what mass of AgCl(s) precipitates out, and what is [Ag +]? Assume no volume change.

AgI: Ksp = 1.5 X 10 -16
AgBr: Ksp = 5.0 X 10 -13
AgCl: Ksp = 1.6 X 10 -10

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