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Rama09 [41]
3 years ago
11

What is the equilibrium constant of pure water at 25°C?

Chemistry
2 answers:
RSB [31]3 years ago
8 0

Answer:

10↑-14

Explanation:

STatiana [176]3 years ago
5 0

Answer: 1.0 × 10-14

Explanation:

Pure water, represented as

H2O --> [H+] + [OH -]

undergoes a reversible reaction in which both H+ and OH- are generated.

The equilibrium constant for this reaction, called the water dissociation constant, Kw, is 1.0 × 10-14 at 25 °C.

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First calculate the mole fraction of each substance:
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Cyclohexane: 1.45 ÷ (2.88 mol + 1.45 mol) = 0.335
Raoult's Law: P(total) = P(acetone) · χ(acetone)  + P(cyclohexane) · χ(cyclohexane).
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