Acid and Bases Lecture #6

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Kylie Joe 2A
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Acid and Bases Lecture #6

Postby Kylie Joe 2A » Sun Jan 17, 2021 12:17 am

In the Lecture 6 example of partial dissociation, we used 1.3x10^-3 as the concentration for CH3COO-, is this because there are the same molar ratio of CH3COO- in the reaction as there is H3O+? If not, how did we figure out that the concentration of CH3COO- was this value?

Bailey Giovanoli 1L
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Re: Acid and Bases Lecture #6

Postby Bailey Giovanoli 1L » Sun Jan 17, 2021 12:22 am

When you do the ICE box, the stoichiometric ratios are what determine this. If you have x of both CH3COO- and H3O+, then they will both have the same concentration.

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Re: Acid and Bases Lecture #6

Postby Hannah_Kim_1I » Sun Jan 17, 2021 4:59 am

If you meant that CH3COO and H30 had the same stoichiometric coefficients, then yes you are correct. We get the value by setting the Ka value equal to [CH3COO][H3O]/[CH3COOH] and plugging in the values from the ICE box to get 1.8 x 10-5 = x2/0.1-x. Then solving for x you get 1.3 x 10-3.

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Re: Acid and Bases Lecture #6

Postby Krish_Ajmani_3J » Sun Jan 17, 2021 10:11 am

The stochiometric coefficients should help determine the ratio of concentrations. For example, if the concentration of a certain molecule is 2X, the amount is doubled.

Bella Bursulaya 3G
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Re: Acid and Bases Lecture #6

Postby Bella Bursulaya 3G » Sun Jan 17, 2021 10:12 am

As others have said, the concentration of x is both the concentration of CH3COO- and H3O+ due to their coefficients. If it was 3 CH3COO-, it would be initial - 3x.

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