Lecture 28 #1

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Claire_Kim_2F
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Lecture 28 #1

Postby Claire_Kim_2F » Wed Dec 09, 2020 12:07 am

I was wondering why HF would stabilize the anion the most and be the strongest acid because I thought Hf would be the weakest because it is the hardest to break and would not disassociate as easily as the other ones listed in the chart?

Lillian Ma 1I
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Re: Lecture 28 #1

Postby Lillian Ma 1I » Wed Dec 09, 2020 12:11 am

Fluorine is the most electronegative which means it pulls the electrons towards it when it is in the compound which in a way means that the H that will be removed contributes less electrons to the compound. This means that the anion left behind is more stable because it was already slightly negative to start with.

Victor Li 2A
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Re: Lecture 28 #1

Postby Victor Li 2A » Wed Dec 09, 2020 12:11 am

I thought the lecture stated that HF is a weak acid since the bond between H and F is very strong and short, thus harder to dissociate and donate H+. In terms of acidic strength: HF < HCl < HBr < HI.

Benjamin_Hugh_3F
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Re: Lecture 28 #1

Postby Benjamin_Hugh_3F » Wed Dec 09, 2020 12:24 am

I am also slightly confused because HF is known to be a weak acid, but sharing my second thought if we look at it from a relation between acid and conjugate base, a stable(weaker) conjugate base means it has a strong acid. If you look at the equation: HF + H20 --> F- + H30+, F- is a stable conjugate base; thus, HF would be a strong acid.

Kiyoka Kim 3C
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Re: Lecture 28 #1

Postby Kiyoka Kim 3C » Wed Dec 09, 2020 6:10 am

I think the example is showing how the fluorine atom in the acid (not HF) is stabilizing the anion because it has high electron withdrawing ability.

Alexandra Salata 2L
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Re: Lecture 28 #1

Postby Alexandra Salata 2L » Thu Dec 10, 2020 1:08 am

Bond strength is related to the length of the bond, and because Iodine has a much larger atomic radius than Fluorine, HI has a much longer, and therefore weaker, bond. The hydrogen is removed fairly easily, making HI a stronger acid


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