Let solubility of AgCl is 'S'
∴ KspAgCl=S2
S=√Ksp=10−5
When 100mL of both AgCl and NaBr are mixed concentration of Ag+ and Br− will be:
[Ag+]=100×10−5200=0.5×10−5M
[Br−]=100×0.03200=0.015M
Ionic product of AgBr=0.5×10−5×0.015=7.5×10−8
Since, ionic product of AgBr is greater than its solublity product, hence AgBr will be precipitated.