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

Question

A 165.0 mL165.0 mL solution of 2.418 M2.418 M strontium nitrate is mixed with 205.0 mL205.0 mL of a 2.465 M2.465 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate.

mass: 31.739

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+]= 1.366

[NO−3]=

[Sr2+]=

[F−]= 0

Answer 1

[Sr²⁺] = 0.395 M  |  [F⁻] = 0 M  |  [Na⁺] = 1.37 M  |  [NO₃⁻] = 2.16 M

Your 31.739 g for the precipitate is right, and it confirms the limiting reactant — so the hard part is already done. Here is the rest.

Step 1 — moles of every ion before mixing

Sr(NO₃)₂: (0.1650 L)(2.418 M) = 0.3990 mol, giving 0.3990 mol Sr²⁺ and 0.7979 mol NO₃⁻ (two nitrates per formula unit — easy to miss).

NaF: (0.2050 L)(2.465 M) = 0.5053 mol, giving 0.5053 mol Na⁺ and 0.5053 mol F⁻.

Step 2 — the reaction

Sr²⁺(aq) + 2 F⁻(aq) → SrF₂(s)

Consuming all 0.3990 mol Sr²⁺ would take 0.7979 mol F⁻, and only 0.5053 mol is present. F⁻ is limiting.

F⁻ consumed: all 0.5053 mol
Sr²⁺ consumed: 0.5053 ÷ 2 = 0.2527 mol
Sr²⁺ left over: 0.3990 − 0.2527 = 0.1463 mol

Step 3 — the total volume is what you divide by

V_total = 165.0 + 205.0 = 370.0 mL = 0.3700 L

This is the step most people miss. Every surviving ion is now diluted into the combined volume, not the volume of the solution it arrived in.

Step 4 — final concentrations

[Sr²⁺] = 0.1463 mol / 0.3700 L = 0.395 M
[F⁻] = 0 M — entirely consumed as the limiting reactant
[Na⁺] = 0.5053 mol / 0.3700 L = 1.37 M
[NO₃⁻] = 0.7979 mol / 0.3700 L = 2.16 M

Na⁺ and NO₃⁻ are spectator ions. They take no part in the precipitation, so their moles are unchanged — but their concentrations still drop, because the volume grew. Reporting their original molarities is the second most common error here.

Check it with charge balance

Total positive: 2(0.395) + 1.37 = 2.16 M
Total negative: 2.16 + 0 = 2.16 M ✓

Any solution must be electrically neutral, so this is a genuine check on all four numbers at once — well worth thirty seconds.

About "assuming complete precipitation": that phrase tells you to take [F⁻] as exactly 0 and ignore K_sp. In reality SrF₂ is slightly soluble (K_sp ≈ 4.3 × 10⁻⁹), so a trace of fluoride really does remain — but it is negligible next to 0.395 M Sr²⁺, and the problem is explicitly telling you not to bother with it.