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At a constant 100 C and 1 ATM ambient pressure, I believe it should be 1 as the gaseous and liquid phases would be equally favored. This is due to the fact that the vapor pressure of liquid water at 100 C is 1 ATM.

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13y ago
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7mo ago

The equilibrium constant for water, Kw, is equal to 1.0 x 10^-14 at 25°C. This value represents the product of the concentrations of H+ and OH- ions in pure water when it is at equilibrium. Kw is a measure of the ionization of water molecules into hydronium and hydroxide ions.

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15y ago

1x10^-14

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Q: Equilibrium constant for water Kw
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What is the difference between Kc and Kw for water?

Kc is the equilibrium constant for a chemical reaction involving water, whereas Kw is the equilibrium constant for the autoionization of water to form hydronium and hydroxide ions. Kw has a fixed value at a given temperature (1.0 x 10^-14 at 25°C), while Kc can vary depending on the specific chemical reaction.


How is the Kw constant developed?

The Kw constant is derived from the auto-ionization of water, where water molecules can transfer a proton to each other to form hydronium and hydroxide ions. The equilibrium constant for this reaction is the Kw constant, which is the product of the concentrations of hydronium and hydroxide ions in water at a given temperature.


What is the only way to change the value of Kw?

The value of Kw, which is the equilibrium constant for the autoionization of water, can be changed by changing the temperature of the water. As temperature increases, the value of Kw also increases because the ionization of water is an endothermic process.


What does K subscript w mean?

Kw is the symbol for the equilibrium constant of water, which represents the auto-ionization of water into hydrogen ions and hydroxide ions. Its value under standard conditions is 1.0 x 10^-14 at 25°C.


How is Kw valid?

Kw is the ionisation constant for water at 25°C which value is 1.0x10^-14. (chemistry)In water at any pH the equilibrium state Kw is defined by and equal to the 'ion product':Kw = [H3O+]*[OH-] = 1.0*10-14at room temperature 25°C


What is the dissociation constant k of pure water?

The dissociation constant (Kw) of pure water is approximately 1 x 10^-14 at 25°C. This value represents the equilibrium constant for the autoionization of water into H+ and OH- ions.


What is ion product constant?

Ion product constant is the product of the concentrations of the ions in a solution at equilibrium. In water, the ion product constant for pure water is Kw = [H3O+][OH-] = 1.0 x 10^-14 at 25°C. It is used to calculate the pH of a solution and can be used to determine if a solution is acidic, neutral, or basic.


What is meant by ionic product of water?

The ionic product of water refers to the equilibrium constant for the dissociation of water into its ions, H+ and OH-. It is represented by the equation: Kw = [H+][OH-]. At 25°C, the value of Kw is 1.0 x 10^-14.


Does taking water out of an equilibrium reaction change the equilibrium constant?

No, removing water from an equilibrium reaction does not change the equilibrium constant. The equilibrium constant is determined by the stoichiometry of the reaction and temperature, not by the presence or absence of water.


What is the value for the ionization constant Kw for water at 25 degrees celsius?

The ionization constant Kw for water at 25 degrees Celsius is 1.0 x 10^-14.


Why kw of water remains same when small amount of acid or base is added?

Yes, it is an equilibrium constant, ONLY depending on the kind of reaction (Kw for water protolysis) and on temperature (according to Arrhenius) and never, NEVER on concentrations of its reactants and products:that is just why it is called a 'C O N S T A N T'


Does Kw show the interdependence of the H3O plus and OH- in aqueous solutions?

Yes, the equilibrium constant for water, Kw, shows the interdependence of H3O+ and OH- in aqueous solutions. It represents the auto-ionization of water into H3O+ and OH- ions and helps quantify the balance between acidic and basic conditions in a solution. At 25°C, Kw is equal to 1.0 x 10^-14 mol2/L2.