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Topic: Equilibrium Equations HW  (Read 1679 times)

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Offline sleepis4theweak

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Equilibrium Equations HW
« on: March 16, 2018, 12:43:19 AM »
For the equation HF(aq) + H2O(l) <-> H3O+(aq) + F-(aq), the Ka is 3.5 * 10-4. If we put 1.2 mol of HF in 2.00 L of water, what is the molarity of the H3O+?

The thing that confuses me about this question is that I can't add liquids to the equilibrium equation so I'm not exactly sure how to go about this.
« Last Edit: March 16, 2018, 01:02:25 AM by sleepis4theweak »
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Offline chenbeier

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Re: Equilibrium Equations HW
« Reply #1 on: March 16, 2018, 01:34:32 AM »
You can write the equation without the water.
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Offline XeLa.

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Re: Equilibrium Equations HW
« Reply #2 on: March 16, 2018, 09:32:24 PM »
Usually with acid-base problems, we ignore [H2O] in the equilibrium equation. The only reason why they mentioned volume of water, I assume, is because they want you to calculate initial molarity of HF. Do that and apply the process you would usually use to solve basic pH problems.
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