Starting with 10.3 ml of 3.40 × 10−2 M BaCl2, how many grams of ZnSO4 can be added to the solution?
Starting with 10.3 ml of 3.40 × 10−2 M BaCl2, how many grams of ZnSO4 can be added to the solution before BaSO4 begins to precipitate ?
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Ksp (BaSO4)= 1.5 10^-9=(Ba2+)(SO42-)=3.4 10^-2(SO42-)
(SO42-)=4.4 10^-8 M
MM(ZnSO4)= 161.4 g/mol
to get grams in 10.3 mL =0.0103 L
0.0103 L(161.4 g/mol)(4.4 10^-8 mol/L)= 7.3 10^-8 g