Concentration of SO42− in BaSO4 solution M1V1=M2V2 1×50=M2×500 M2=
1
10
For just precipitation Ionic product =Ksp [Ba2+][SO42]=Ksp(BaSO4) [Ba2+]×
1
10
=10−10 [Ba2+]=10−9M in 500mL solution Thus [Ba2+ ] in original solution (500−50=450mL) ⇒M1×450=10−9×500 [ where M1= Molarity of original solution] M1=