A 165.0 ml solution of 2.724M strontium nitrate is mixed with 215.0ml of a 3.120M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate.

Question

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

Answer 1

To tackle this stoichiometry question, let's first write the complete ionic and net ionic equations for what happens when these two solutions are mixed:

(1) Sr²⁺ (aq) + 2NO₃^- (aq) + 2Na⁺ (aq) + 2F^- (aq) → SrF₂ (s) + 2Na⁺ (aq) + 2NO₃^- (aq)

(2) Sr²⁺ (aq) + 2F^- (aq) → SrF₂ (s)

As seen in the net ionic equation (2), the meaningful chemistry that occurs is the precipitation of Strontium Fluoride from solution.

Assuming complete precipitation, the mass of SrF₂ (s) made can be found by completing the following stoichiometric steps:

2.724M Sr(NO₃)₂ x .165L = 0.4495 mol Sr²⁺

3.120M NaF x .215L = 0.6708 mol F^-

The limiting reactant is fluorine since 0.4495 mol Sr²⁺ would make 0.4495 mol SrF₂ while 0.6708 mol F^- can only make 0.3354 mol SrF₂ (because two fluorine atoms are found in one molecule of SrF₂).

Thus, 0.3354 mol SrF₂ will be made, which when multiplied by 125.62g/mol, the molar mass of SrF₂, gives 42.13g SrF₂.

Let's now calculate the final concentrations of each ion after the precipitation reaction is completed.

2.724M Sr(NO₃)₂ x .165L = 0.4495 mol Sr(NO₃)₂ = 0.8989 mol NO₃^-

0.8989 mol NO₃^-/ (.165L + .215L) = 2.366M NO₃^-

3.120M NaF x .215L = 0.6708 mol NaF = 0.6708 mol Na⁺

0.6708 mol Na⁺/ (.165L + .215L) = 1.765M Na⁺

2.724M Sr(NO₃)₂ x .165L = 0.4495 mol Sr(NO₃)₂ = 0.4495 mol Sr²⁺

Since 0.3354 mol Sr²⁺ is used up in the precipitation of SrF₂, 0.1141 mol Sr²⁺ remains in solution.

0.1141 mol Sr²⁺/(.165L + .215L) = 0.3003M Sr²⁺

Since all of the fluoride is used up in the precipitation of SrF₂, there is 0M F^- in solution.