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Oliga [24]
3 years ago
7

When 18.0 mL of a 8.43×10-4 M cobalt(II) fluoride solution is combined with 22.0 mL of a 9.72×10-4 M sodium hydroxide solution d

oes a precipitate form? no (yes or no) For these conditions the Reaction Quotient, Q, is equal to .
Chemistry
1 answer:
Eva8 [605]3 years ago
8 0

Answer:

Q = 1.08x10⁻¹⁰

Yes, precipitate is formed.

Explanation:

The reaction of CoF₂ with NaOH is:

CoF₂(aq) + 2 NaOH(aq) ⇄ Co(OH)₂(s) + 2 NaF(aq).

The solubility product of the precipitate produced, Co(OH)₂, is:

Co(OH)₂(s) ⇄ Co²⁺(aq) + 2OH⁻(aq)

And Ksp is:

Ksp = 3x10⁻¹⁶= [Co²⁺][OH⁻]²

Molar concentration of both ions is:

[Co²⁺] = 0.018Lₓ (8.43x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>3.79x10⁻⁴M</em>

[OH⁻] = 0.022Lₓ (9.72x10⁻⁴mol / L) / (0.018 + 0.022)L = <em>5.35x10⁻⁴M</em>

Reaction quotient under these concentrations is:

Q = [3.79x10⁻⁴M] [5.35x10⁻⁴M]²

<em>Q = 1.08x10⁻¹⁰</em>

As Q > Ksp, <em>the equilibrium will shift to the left producing Co(OH)₂(s) </em>the precipitate

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<h3>What is the rate law of a reaction?</h3>

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The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

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Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

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Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

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The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

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