Le Chatlier's Principle Question

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herewego

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What will occur if the basicity of a saturated solution of Na2SO4 is increased?

I reasoned:

SO4 + H2O -> OH + HSO4

So if you increase OH, you increase the right hand side of the equation. Why wouldn't more Na2SO4 dissolve?

The answer states the Na2SO4 will precipitate out of solution.

Any help would be appreciated.

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If the right hand side increases, it means the reaction Eq will shift to the left...which means more SO4(2-) will be make..therefore NaSO4 will precipitate out of solution

matth87

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If the solution is saturated it is impossible to dissolve any more NaSO4. Increasing -OH will shift the Equilibrium to the left. NaSO4 will increase and since it cant dissolve it will precipitate out.

herewego

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Ok that makes sense. Thanks. I guess I didn't take into consideration the fact that it was saturated. The explanation Kaplan gives doesn't use that in their reasoning either.

Last edited:

Pulsar

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I was thinking SO4(2-) is the conjugate base of a weak acid, so a solution of SO4(2-) will be basic because SO4(2-) + H20 --> HSO4(-) + OH(-). Increasing the OH concentration makes more SO4(2-) and increasing SO4(2-) means more Na2SO4, because Na2SO4 --> 2Na+ + SO4(2-)

housemd?

USAF
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I was thinking SO4(2-) is the conjugate base of a weak acid, so a solution of SO4(2-) will be basic because SO4(2-) + H20 --> HSO4(-) + OH(-). Increasing the OH concentration makes more SO4(2-) and increasing SO4(2-) means more Na2SO4, because Na2SO4 --> 2Na+ + SO4(2-)

H2SO4 is a weak acid???

G1SG2

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I was thinking SO4(2-) is the conjugate base of a weak acid, so a solution of SO4(2-) will be basic because SO4(2-) + H20 --> HSO4(-) + OH(-). Increasing the OH concentration makes more SO4(2-) and increasing SO4(2-) means more Na2SO4, because Na2SO4 --> 2Na+ + SO4(2-)

H2SO4 is a strong acid!!!

wanderer

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Pulsar is correct. HSO4- the conjugate acid of SO4 2- is a weak acid.

Pulsar

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H2SO4 is a strong acid!!!

H2SO4 is strong, but HSO4- is weak. Sulfuric acid is a diprotic acid, but is only strong in it's first ionization.