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Anna007 [38]
2 years ago
15

Acetic acid has the formula ch3cooh. At low ph this protonated, uncharged form is the most abundant form in solution. As the ph

increases, more of the molecule becomes negatively charged acetate: ch3coo–. At which ph will more of the molecule cross the membrane?
Chemistry
1 answer:
slamgirl [31]2 years ago
5 0

Acetic acid is a weak acid. So, the undissociated acid molecules exist in equilibrium with the dissociated form. The equilibrium representing the weak acidic nature of acetic acid is,

CH_{3}COOH(aq)+H_{2}O(l)CH_{3}COO^{-}(aq)+H_{3}O^{+}(aq)

The interior of the cell membrane is hydrophobic. So, the membrane allows uncharged species (hydrophobic) to pass through it. Charged groups cannot easily pass through the membrane. We can say acetic acid can easily pass through the membrane when compared to the charged form acetate ion. At low pH, more of the molecule will cross the membrane as most of the acetic acid will be in undissociated and uncharged form at a lower pH.

pK_{a}of acetic acid is around 4.8. So at a pH value less than 4.8, we can say the more of the acetic acid molecules can cross the membrane.


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Which part of experimental design is most important to a scientist when replicating an experiment? Having exactly the same data
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Answer:

The answer to your question is below:

Explanation:

Having exactly the same data as the previous experiment I think that having the same data as the previous experiment is extremely important but not the most important, for me is the second most important.

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7. You are about to perform some intricate electrical studies on single skeletal muscle fibers from a gastronemius muscle. But f
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4 0
2 years ago
Oxalic acid is a diprotic acid. calculate the percent of oxalic acid (h2c2o4) in a solid given that a 0.7984-g sample of that so
vlada-n [284]
This method of quantitative determination of percent purity is titrimetric reactions. These reactions most commonly involve neutralization reactions between an acid and a base. Then, we look at the neutralization reaction:

H₂C₂O₄ + 2 NaOH ⇒ Na₂C₂O₄ + 2 H₂O

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(0.2283 mol/L NaOH * 0.3798 L * 1 mol H₂C₂O₄/ 2mol NaOH * 90 g/mol H₂C₂O₄) ÷ 0.7984 g *100%
= 488%

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