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Question

a 190ml solution of 2.104M strontium nitrate is mixed with 205ml of a 2.597M sodium fluoride solution. calculate the mass of the resulting strontium flouride precipitate.

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

Na

NO3

Sr

F

Answer 1

First, write a correctly balanced equation for the reaction taking place:

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

Next, find the limiting reactant if there is one:

moles Sr(NO₃)₂ present = 190 ml x 1 L/1000 ml x 2.104 mol/L = 0.39976 moles

moles NaF present = 205 ml x 1 L/1000 ml x 2.597 mol/L = 0.5324 moles

From the balanced eq., it takes 2 mol NaF for each 1 mol Sr(NO₃)₂, NaF is limiting.

Now that we know the limiting reactant, we can find the mass of precipitate, SrF₂.

0.5324 mol NaF x 1 mol SrF₂ / 2 mol NaF = 0.2662 moles SrF₂

mass of SrF₂ formed = 0.2662 moles SrF₂ x 125.62 g/mol = 33.44 g = 33 g (2 sig figs based on 190 ml)

Assuming complete precipitation of SrF₂, we can find the final concentration of the various ions.

The final volume = 190 ml + 205 ml = 395 mls = 0.395 L

Initial moles Sr(NO₃)₂ = 0.3998 moles.

Moles Sr that reacted/precipitated = 0.2662 moles

Moles Sr²⁺ remaining = 0.3998 - 0.2662 = 0.1336 moles Sr²⁺

Final [Sr^2+] = 0.1336 moles/0.395 L = 0.34 M Sr²⁺ (2 sig. figs)

Initital moles NaF2 present = 0.5324 moles

Moles Na+ precipitated = 0

Final [Na+] = 0.5324 mol/0.395 L = 1.3 M Na+

Moles F- precipitated = 0.2662 x 2 = 0.5324 moles

Final [F-] = 0 M F^-

Initial moles NO₃^- = 0.3998 x 2 = 0.7996 moles

Moles precipitated = 0

Final [NO₃^-] = 0.7996 mol/0.395 L = 2.0 M NO₃^-

SUMMARY

Na = 1.3 M

NO₃ = 2.0 M

Sr = 0.34 M

F = 0