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Anna007 [38]
3 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]3 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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4 years ago
Calculate the hight of a column of liquid glycerol in meters required to exert the same pressure as 3.02 m if CCl4?
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Approximately 3.81\; \rm m.

Explanation:

Look up the density \rho of carbon tetrachloride, \rm CCl_4, and glycerol:

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  • Density of glycerol: approximately 1.26\times 10^{3}\; \rm kg \cdot m^{-3}.

Let g denote the gravitational field strength. (Typically g \approx 9.81\; \rm N \cdot kg^{-1} near the surface of the earth.) For a column of liquid with a height of h, if the density of the liquid is \rho, the pressure at the bottom of the column would be:

P = \rho\cdot g \cdot h.

The pressure at the bottom of this carbon tetrachloride column would be:

\begin{aligned} P &= \rho \cdot g \cdot h \\ & \approx 1.59\times 10^{3} \; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1} \times 3.02 \; \rm m \approx 4.71 \times 10^{4} \; \rm N \cdot m^{-2} \end{aligned}.

Rearrange the equation P = \rho\cdot g \cdot h for h:

\displaystyle h = \frac{P}{\rho \cdot g}.

Apply this equation to calculate the height of the liquid glycerol column:

\begin{aligned}h &= \frac{P}{\rho \cdot g} \\ &\approx \frac{4.71 \times 10^{4}\; \rm N \cdot m^{-2}}{1.26 \times 10^{3}\; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1}} \approx 3.81\; \rm m\end{aligned}.

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3 years ago
Examine the general structure of a molecule.
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Answer:

Explanation:

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3 years ago
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