Question
Thu March 24, 2011 By: Kuldeep Shrivastava
determine the solubility of barium sulphate in 0.05M in barium chloride solution.(Ksp-BaSo4=1.1*

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Expert Reply
Thu April 07, 2011
BaCl2→ Ba2+ +SO42-
(Ba2+)  =.05M = 5 X 10-2 M
(SO42-)  =1.1 X 10 -10  /  5 X 10-2
                 = .22 X 10 -8 mol l-1
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