A 165.0 mL solution of 2.282 M strontium nitrate is mixed with 220.0 mL of a 2.278 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate.

Question

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

Find

Na+

NO-3

Sr2

F-

Answer 1

Write the correctly balanced equation for the reaction:

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

moles Sr(NO₃)₂ present = 165.0 ml x 1 L / 1000 ml x 2.282 mol / L = 0.3765 mols Sr(NO₃)₂

moles NaF present = 220.0 ml x 1 L / 1000 ml x 2.278 mol / L = 0.5012 mols NaF

Because the mol ratio in the balanced equation shows that we need twice as much NaF as Sr(NO₃)₂

and we don't have twice as much, this means NaF is going to be LIMITING, and production of SrF₂

precipitate will be determined by amount of NaF present, with Sr(NO₃)₂ being left over in excess.

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

The only ions that are removed from solution are the Sr²⁺ and F^- ions, as they are incorporated in the ppt.

So, if we calculate initial moles of all the ions, and subtract the mols of Sr²⁺ and F^- in the ppt, we will end up with the final moles of each ion in solution. Divide this by final volume to final concentrations.

Initial mols Sr²⁺ = 0.3765 mols

Initial mols NO₃^- = 2 x 0.3765 = 0.7530 mols

Initial mols Na⁺ = 0.5012 mols

Initial mols F^- = 0.5012 mols

Final mols Sr²⁺ = 0.3765 mols - 0.2506 mols = 0.1259 mols Sr²⁺ left

Final mols NO₃^- = 0.7530 mols NO₃^-

Final mols Na⁺ = 0.5012 mols Na⁺

Final mols F^- = 0.5012 mols - 0.5012 mols = 0 mols F^-

Final volume = 165 ml + 220 ml = 385 mls = 0.385 M

Final concentrations After the reaction:

[Sr²⁺] = 0.1259 mol / 0.385 L = 0.3270 M

[NO₃^-] = 0.7530 mol / 0.385 L = 1.956 M

[Na⁺] = 0.5012 mol / 0.385 L = 1.302 M

[F^-] = 0

Answer 2

Upvote my answer and if u have more questions ask direct [email protected] exam help quiz etc

Answer: To find the mass of the resulting strontium fluoride precipitate, we need to first determine which ion is the limiting reagent in the reaction.

The balanced chemical equation for the reaction between strontium nitrate and sodium fluoride is:

Sr(NO3)2(aq) + 2 NaF(aq) → SrF2(s) + 2 NaNO3(aq)

Using the volumes and molarities provided, we can calculate the number of moles of each reactant:

moles of Sr(NO3)2 = (165.0 mL)(2.282 mol/L) = 376.83 mmol

moles of NaF = (220.0 mL)(2.278 mol/L) = 501.16 mmol

Since the stoichiometric ratio of Sr(NO3)2 to NaF is 1:2, the limiting reagent is Sr(NO3)2, which means that all of the Sr(NO3)2 will react with 2 times the amount of NaF to form SrF2.

moles of Sr(NO3)2 that react = 376.83 mmol

moles of NaF that react = 2(376.83 mmol) = 753.66 mmol

The mass of SrF2 that forms can be calculated from the moles of Sr(NO3)2 that react, using the molar mass of SrF2:

moles of SrF2 formed = 376.83 mmol

mass of SrF2 formed = (mol of SrF2)(molar mass of SrF2) = (376.83 mmol)(125.62 g/mol) = 47.38 g

Therefore, the mass of the resulting strontium fluoride precipitate is 47.38 g.

To calculate the final concentration of each ion, we need to first determine the total volume of the resulting solution.

Total volume of resulting solution = 165.0 mL + 220.0 mL = 385.0 mL

To calculate the final concentration of Na+ and NO3-, we can use the balanced chemical equation to determine that they will both have a concentration of 2.282 M in the resulting solution, since they are both produced in equal amounts as the reactant Sr(NO3)2.

[Na+] = 2.282 M

[NO3-] = 2.282 M

To calculate the final concentration of Sr2+ and F-, we need to use the stoichiometry of the reaction to determine how much of each ion remains in solution after the precipitation of SrF2.

moles of Sr(NO3)2 that react = 376.83 mmol

moles of SrF2 formed = 376.83 mmol

moles of Sr2+ remaining = 0 mmol (all reacted)

moles of NaNO3 formed = 753.66 mmol

moles of F- remaining = 753.66 mmol - 2(376.83 mmol) = 0 mmol (all reacted)

Therefore, the final concentration of Sr2+ is 0 M, and the final concentration of F- is 0 M.