Chemistry, 29.07.2019 13:30, Cheflulu5727

This solution contains equal concentrations of both hf(aq) and naf(aq). write the proton transfer equilibrium (net ionic) reaction that includes f– as a reactant. include states of matter.

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We are asked to provide the net ionic equation for the reaction of HF (aq) and NaF (aq). HF is a weak acid and is in the following equilibrium:

HF (aq) ⇄ H⁺ (aq) + F⁻ (aq)

Meanwhile, NaF (aq) is an ionic compound that will dissociate completely in aqueous solutions:

NaF (aq) → Na⁺ (aq) + F⁻ (aq)

We can combine the ionic species with HF, as we are told to show F⁻ as a reactant:

HF (aq) + Na⁺ (aq) + F⁻ (aq) → HF (aq) + Na⁺ (aq) + F⁻ (aq)

We can eliminate the spectator ions, which in this case are Na⁺ ions, and this leaves us with the net ionic equation involving F⁻:

HF (aq) + F⁻ (aq) → HF (aq) + F⁻ (aq)

In this instance, the proton is just transferred between F⁻ ions and the end result is the formation of more HF, so there is no net reaction taking place.

answered: Guest

Where is the following?

answered: Guest

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.