Chemistry, 18.07.2020 02:01 Dragonfierz666

A solution contains A13+ and Co2+. The addition of 0.3731 L of 1.735 M NaOH results in the complete precipitation of the ions as Al(OH), and Co(OH)2. The total mass of the precipitate is 22.73 g. Find the masses of Al3+ and Co2+ in the solution.

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Answer from: QuarkyFermion

The correct answer is mass of Al3+ will be 3.23 grams and the mass of Co2+ will be 8.50 grams.

Explanation:

Based on the given information, 0.3731 L of 1.735 M of NaOH is added in a solution resulting in the precipitation of the ions as Al(OH)₃ and Co(OH)₂. Thus, the moles of NaOH will be molarity × V(L) = 1.735 × 0.3731 L = 0.647 moles.

The mass of the precipitate given is 22.73 grams.

Now let us assume that the mass of Al(OH)₃ will be x grams and the mass of Co(OH)₂ will be (22.73-x) grams

Therefore, the moles of Al(OH)₃ will be x grams/78 g/mol and as 3OH⁻ ions are needed so the moles will be 3x/78 mole.

And, the moles of Co(OH)₂ will be (22.73-x)grams/92.94 g/mol and as 2OH⁻ ions are needed so the moles will be 45.46-2x/92.94 moles.

Now the equation will become,

3x/78 + 45.46-2x/92.94 = 0.647 moles

0.03846 x + 0.489 - 0.02152 x = 0.647

0.01694 x + 0.489 = 0.647

0.01694 x = 0.158

x = 0.158/0.01694

x = 9.327 grams

Hence, the mass of Al(OH)₃ is 9.327 grams, and the mass of Al³⁺ will be,

= 9.327 gm/78 g/mol × 27 g/mol = 3.23 grams

Now the mass of Co(OH)₂ will be, (22.73 - 9.327) grams = 13.403 grams

the mass of Co²⁺ will be,

= 13.403 grams / 92.94 g/mol × 58.94 g/mol = 8.50 grams

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.

Answer from: Quest

08.01 hope this and can i get brainlist? ?

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A solution contains A13+ and Co2+. The addition of 0.3731 L of 1.735 M NaOH results in the complete...