Answer to Question #62284 in General Chemistry for Will Benedict

Question

What volume of .100 M NaOH is required to precipitate all of the Ni^2+ ion from 150.0 mL of .250 M Ni(NO2)2?

What mass of BaSO4 would precipitate when 100.0 mL of a .200 M BaCl2 solution is mixed with 100.0 mL of .100 M Fe2(SO4)3 solution?

Answer 1

The amount of Ni2+ in Ni(NO2)2 solution is 0.250 mol/L*0.150 L = 0.0375 mol. It requires twice more NaOH; 0.0375 mol*2 = 0.075 mol or 0.075 mol/0.100 mol/L = 0.75 L (750 mL).
100.0 mL of 0.200 M BaCl2 contains 0.02 mol of Ba2+ and 100.0 mL of 0.100 M Fe2(SO4)3 contains 0.03 mol of SO42-. Thus, the amount of BaSO4 is 0.02 mol or 4.66 g.