Chemistry, 19.04.2020 06:26 TGJuuzouSuzuya

Write the balanced NET ionic equation for the reaction when AlCl₃ and NaOH are mixed in aqueous solution. If no reaction occurs, write only NR.

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Answer from: allakhalilpea0zc

The balanced NET ionic equation for the reaction when AlCl₃ and NaOH are mixed in an aqueous solution is:

Al³⁺(aq) + 3OH⁻(aq) → Al(OH)₃(s)

The reaction between AlCl₃ and NaOH is an exchange reaction:

AlCl₃(aq) + 3NaOH(aq) → Al(OH)₃(s) + 3NaCl(aq) (1)

In this reaction, the ions Cl⁻ and OH⁻ are exchanged between the ions Al⁺³ and Na⁺.

To write the NET ionic equation, we need to write the total ionic equation:

Al³⁺(aq) + 3Cl⁻(aq) + 3Na⁺(aq) + 3OH⁻(aq) → Al(OH)₃(s) + 3Na⁺(aq) + 3Cl⁻(aq) (2)

Since the Al(OH)₃ precipitates, we cannot write it as an ion because it does not dissociate in the aqueous solution.

The net ionic equation of reaction (2) is the following:

Al³⁺(aq) + 3OH⁻(aq) → Al(OH)₃(s) (3)

All repeated ions in reactants and products do not participate in the net ionic equation. That is why we canceled the Na⁺ and Cl⁻ ions in the net ionic equation (eq 3).

Therefore, the balanced net ionic equation is:

Al³⁺(aq) + 3OH⁻(aq) → Al(OH)₃(s)

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Answer from: mzbugsbunny3029

$Al^{3+} + 3OH^{-} --- Al(OH)_{3}$

Explanation:

AlCl3 + 3NaOH ---> Al(OH)3 + 3NaCl

$Al^{3+} +3Cl^{-} +3Na^{+}+3OH^{-}---AL(OH)_{3}+3Na^{+}+3OH^{-}\\\\Al^{3+} + 3OH^{-} --- Al(OH)_{3}$

Write the balanced NET ionic equation for the reaction when AlCl₃ and NaOH are mixed in aqueous solu

Answer from: Quest

Okay this will take 1 hour okay dont forget me.

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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