Question about strontium sulfate

Question

Strontium sulfate is an insoluble solid. Determine the molar concentration of sulfate ions in a solution made by mixing 183 mL of a 1.95 mol/L solution of Strontium nitrate with 225 mL of a 1.84 mol/L solution of Lithium sulfate

Answer 1

Hey, Karen, it's the same kind of problem again...

 

Write the chemical equation:

 

Sr(NO3)2 + Li2SO4 → SrSO4 + LiNO3

 

Balance:

 

Sr(NO3)2 + Li2SO4 → SrSO4 + 2LiNO3

 

.183L Sr(NO3)2 x 1.95 mole Sr(NO3)2/L = 0.357 mole Sr(NO3)2

 

.225 L Li2SO4 x 1.84 mole Li2SO4/L = 0.414 mole Li2SO4

 

For every SrSO4, which precipitates out, you use one SO4, so all 0.357 mole of Sr(NO3)2 will be used up and will make 0.357 moles of SrSO4.

 

If you begin with 0.414 mole Li2SO4 and you use up 0.357 moles of the SO4, then, .0570 moles of SO4 2- will remain in solution.

 

.0570 moles of SO4 2-/.183L + .225 L  ( this is the sum of the volumes of the solutions = .408L) = .140 moles SO42-/L