Fundementals J15 NaC6H5O is acid/base?

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Madeline Offerman 3G
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Fundementals J15 NaC6H5O is acid/base?

Postby Madeline Offerman 3G » Sun Nov 29, 2015 11:53 am

In this problem you are given NaC6H5O and are asked to write the equation for the proton transfer with water. How do you know that this compound is a base and should gain an H+ proton?
Last edited by Madeline Offerman 3G on Sun Nov 29, 2015 10:50 pm, edited 1 time in total.

Julia Spaczai 2G
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Re: Fundementals J15 NaC6H5O is acid/base?

Postby Julia Spaczai 2G » Sun Nov 29, 2015 12:46 pm

In the question, it states that either the cation or anion of the salt is a weak acid or a weak base. The anion of NaC6H5O is C6H5O-, which is a weak acid. You use this anion when writing the equation for the proton transfer with water.

BriannaWillingham_1G
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Re: Fundementals J15 NaC6H5O is acid/base?

Postby BriannaWillingham_1G » Sun Nov 29, 2015 5:55 pm

How is C6H5O- a weak acid?
Table 12.1 on page 477 in chapter 12 shows that phenol, c6h5oh is a weak acid, so shouldn't it's conjugate base, c6h5o-, be a strong base??? (This is due to the conjugate seesaw concept which states that if an acid or base is strong, then it's conjugate is weak and vice versa)

Also, if you look at its structure, doesn't that negative sign indicate that it really wants a proton? (And thus is a strong base?)


The solution manual also shows that c6h5o- is in fact a base.

Anne Cam 3A
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Re: Fundementals J15 NaC6H5O is acid/base?

Postby Anne Cam 3A » Sun Nov 29, 2015 6:41 pm

C6H5O- is a base because it accepts protons, not donates them. Since phenol is a weak acid, it would not produce C6H5O- as easily, and its dissociation reaction requires an equilibrium symbol:

C6H5OH(aq) + H2O(l) <--> C6H5O-(aq) + H3O+(aq)

C6H5O- would also be readily converted back to C6H5OH, making it a weak base.

Matthew Mar 1J
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Re: Fundementals J15 NaC6H5O is acid/base?

Postby Matthew Mar 1J » Thu Dec 06, 2018 7:06 pm

By equilibrium symbol, are you referring to the notation used in class with two arrows stacked on top of each other? What exactly does this symbol mean, is it just there to indicate that the reaction is prone to go either way?


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