(SO2) 6.0×104×40.0 48×10-4mol.L 50.0 C(Ba+) 0.010×10.0 =2.0×10-3mol·L 50.0 J={Cc(SO4){c0(Ba2+)} 4.8×104×20×10 机邵学电國图 9.6×10 sp=1.1×1040 J>K,所以有BaSO沉淀析出。4 1 4 2 o 4 4.8 10 mol L 50.0 6.0 10 40.0 (SO ) - - - - =     c = 7 9.6 10- =  4 3 4.8 10 2.0 10 - - =    { (SO )}{ (Ba )} 2 o 2 o 4 - + J = c c 2 3 1 o 2.0 10 mol L 50.0 0.010 10.0 (Ba ) + - - =    c = 10 sp 1.1 10- K =  J > Ksp ，所以有BaSO4 沉淀析出