What mass of solid manganese chloride must be added to cause a precipitate to begin?

Manganese sulfide (MnS) has Ksp of 5 x 10^-14. What mass of solid manganese chloride MnCl2 would have to be added to 100 mL or a 0.025 M Na2S solution to cause precipitation of MnS to begin?





I'm stuck on this question and I need the help :( any is appreciated!|||Ksp = [Mn] [S]





5 e-14 = [Mn] [0.025M]





Mn needed is = 2.0 e-12 molar





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in 100ml , @ 2.0 e-12 molar .... that's 2.0 e-13 moles of Mn+2








2.0 e-13 moles of Mn+2 is provided by an equal number of moles of MnCl2 = 2.0 e-13 mol of MnCl2








using molar mass


2.0 e-13 mol of MnCl2 @ 125.84 g/mol = 2.52 e-11 grams of MnCl2





your data has only 1 sig fig in the "5 x 10^-14"


rounded to 1 sig fig


rounding off the the answer


2.52 e-11 grams of MnCl2


is your call