Chemistry, 08.07.2020 01:01 martinezizzie

Aqueous solutions of sodium sulfate and potassium chloride are mixed. What is the precipitate and how many molecules are formed?

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Answer from: JoJo3671

The two products of this reaction, Sodium Chloride and Potassium Sulfate, are both soluble in water, hence, there's no precipitate formed from this reaction.

Explanation:

Sodium sulfate is Na₂So₄

Potassium Chloride is KCl

When they both react, theres a double displacement where ions and radicals are exchanged

Na₂SO₄ + KCl → NaCl + K₂SO₄

The products are

NaCl - Sodium Chloride

K₂SO₄ - Potassium Sulfate

The two products are soluble in water, hence, there's no precipitate formed from this reaction.

And we would need numerical values of the number of moles of the two reactants mixed to obtain the number of moles of products formed and then the number of molecules from Avogadro's constant.

Hope this Helps!!!

Answer from: Quest

answeridentity;

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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