A 180.0 mL solution of 2.400 M strontium nitrate is mixed with 200.0 mL of a 3.088 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate.

Question

A 180.0 mL solution of 2.400 M strontium nitrate is mixed with 200.0 mL of a 3.088 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate.

Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration.

[Na+]= ____ M

[NO−3]= ___ M

[Sr2+]= ___ M

[F−]= ___ M

Answer 1

Sr(NO₃)₂(aq) + 2NaF(aq) ==> SrF₂(s) + 2NaNO₃ ... balanced equation

moles Sr(NO3)2 = 0.180 L x 2.400 mol/L = 0.432 moles

moles NaF = 0.200 L x 3.088 mol/L = 0.6176 moles

NaF is LIMITING REACTANT b/c it takes 2 NaF per 1 Sr(NO₃)₂ and there isn't enough NaF.

Mass of Sr(NO₃)₂ precipitate = 0.6176 mol NaF x 1 mol SrF₂ / 2 mol NaF x 125.6 g SrF₂/mol = 38.8 g

Final volume = 180 ml + 200 ml = 380 ml = 0.380 L

[F^-] = 0 since all the F^- is precipitated as SrF₂

[Sr²⁺] = 0.432 mol - 0.309 mol = 0.123 mol/0.380 L = 0.324 M

[Na⁺] = 0.6176 moles / 0.380 L = 1.63 M

[NO₃^-] = 0.432 moles x 2 = 0.864 mol/0.380 L = 2.27 M