Kw in Lecture

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Eileen Quach Dis 2A
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Kw in Lecture

Postby Eileen Quach Dis 2A » Mon Jan 11, 2021 2:10 pm

In today's lecture for the example that involved the reaction Ba(OH)2--> Ba+2OH, I was wondering why Professor Lavelle wrote Kw as equal to the hydronium concentration multiplied by the hydroxide equation. Why didn't he include Ba in the equilibrium constant when it's not a solid or liquid?

Arezo Ahmadi 3J
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Re: Kw in Lecture

Postby Arezo Ahmadi 3J » Mon Jan 11, 2021 2:20 pm

Since Ba(OH)2 is a strong base, Ba2+ is a spectator ion. Spectator ions end up being cancelled out, so that is why it is not included in the Kw equation.

Sophia Hu 1A
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Re: Kw in Lecture

Postby Sophia Hu 1A » Mon Jan 11, 2021 2:22 pm

Kw is the equilibrium constant of water not for the reaction of Ba(OH)2.

What he wanted to show with that chemical reaction equation is that Ba(OH)2 is a strong base which is completely ionized that will form 2 OH- molecules. Since it is completely ionized then the 0.0030 M of Ba(OH)2 will completely dissociate into OH-, but there are 2 moles of OH- which is why it is 0.0060.

Kw is considered separately. He used Kw = [H3O+][OH-] in order to find the concentration of [H3O+] because we know Kw (always constant) and [OH-].

Abby Lam 3F
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Re: Kw in Lecture

Postby Abby Lam 3F » Mon Jan 11, 2021 2:33 pm

Kw is not the equilibrium constant of the Ba equation. Kw is separate and is equal to [H3O][OH]. He isolated [H3O] so he could solve for it.

Alara Aygen 3K
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Re: Kw in Lecture

Postby Alara Aygen 3K » Mon Jan 11, 2021 2:35 pm

Kw is the equilibrium constant for: [H3O+][OH-] (the autoprotolysis of water)

Inderpal Singh 2L
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Re: Kw in Lecture

Postby Inderpal Singh 2L » Mon Jan 11, 2021 2:39 pm

Kw is the equilibrium constant for [H30+][OH-], not the equilibrium constant for the Ba equation.


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