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hodyreva [135]
3 years ago
9

When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate, a precipitation reaction oc

curs. Write the balanced net ionic equation of the reaction. Include charges on the ions, where applicable. Include coefficients only when they are different than ?
Chemistry
1 answer:
Georgia [21]3 years ago
3 0

Answer:

2Cl^{-}_{(aq)}  + 2K^{-} _{(aq)}  = 2KClx_{(aq)}

Explanation:

The reaction is a precipitation reaction. In other words, the two aqueous solutions react to give the solid salt, potassium chloride (KCl) in this case.

SrCl_{2} _{(aq)}  + K_{2}SO_{4} _{(s)} = SrSO_{4} _{(aq)} + 2KCl_{(s)}

Because the reaction product is a solid, the net ionic equation can be written as:

2Cl^{-}_{(aq)}  + 2K^{-} _{(aq)}  = 2KClx_{(aq)}

This is the resultant equation after removing spectator ions: sulfate and strontium.

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Consider the reaction:
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Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

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K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

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Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

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Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

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(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

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