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Stoichiometry and solutionsNational 2018

After mixing 30.0 mL of 0.30 M $\ce{Ca(NO3)2}$ solution and 15.0 mL of 0.60 M $\ce{NaF}$ solution, which ions are present in solution at concentrations of at least 0.10 M?

I. $\ce{Ca^2+}$

II. $\ce{F-}$

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Explanation

draft

Step 1: Find moles of each ion before reaction.

$n(\ce{Ca^2+}) = 0.0300\text{ L} \times 0.30\text{ M} = 9.0\times10^{-3}\text{ mol}$
$n(\ce{F-}) = 0.0150\text{ L} \times 0.60\text{ M} = 9.0\times10^{-3}\text{ mol}$

Step 2: Precipitation reaction.

$\ce{Ca^2+ + 2F- -> CaF2 (s)}$

This requires 2 mol $\ce{F-}$ per mol $\ce{Ca^2+}$ . To consume all the $\ce{Ca^2+}$ (9.0×10⁻³ mol) would need $1.8\times10^{-2}$ mol $\ce{F-}$ , but only $9.0\times10^{-3}$ mol is available. So $\ce{F-}$ is the limiting reagent and is essentially completely consumed.

Step 3: Remaining $\ce{Ca^2+}$ .

Ca²⁺ consumed = half of the F⁻ used = $\tfrac{1}{2}(9.0\times10^{-3}) = 4.5\times10^{-3}$ mol.

Ca²⁺ remaining = $9.0\times10^{-3} - 4.5\times10^{-3} = 4.5\times10^{-3}$ mol.

Step 4: Convert to concentration.

Total volume = 30.0 + 15.0 = 45.0 mL = 0.0450 L.

$[\ce{Ca^2+}] = \frac{4.5\times10^{-3}\text{ mol}}{0.0450\text{ L}} = 0.10\text{ M}$

This just meets the 0.10 M threshold, so statement I is true.

Since $\ce{F-}$ is (essentially) completely precipitated as $\ce{CaF2}$ (which is quite insoluble), its remaining concentration is far below 0.10 M — statement II is false.

Answer: (A) I only.

Correct: F⁻ is limiting, consumed almost completely by precipitation; leftover Ca²⁺ = 4.5×10⁻³ mol / 0.0450 L = 0.10 M, meeting the threshold, while F⁻ concentration is negligible.

This choice results from assuming Ca²⁺ is the limiting reagent (reversing the stoichiometric roles), incorrectly concluding F⁻ remains in excess while Ca²⁺ is fully consumed.

This comes from ignoring the 1:2 stoichiometry of the precipitation reaction and simply diluting both initial concentrations without accounting for any reaction, making both ions appear to remain above 0.10 M.

This results from assuming both ions are completely removed by precipitation (treating the reaction as if it goes to 100% completion for both reactants), rather than recognizing that Ca²⁺ is in excess relative to F⁻.

Concept · Limiting reagent in precipitation reactions (solution stoichiometry)