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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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How can you increase the amount of work done when you lift a book explain
Gnoma [55]
<h2>Work done = mgh </h2>

Explanation:

  • In this case, while lifting the book we are working against the force of gravity.

Using the Newton's laws, we can find the force F required for lifting the book having mass (m) and acceleration due to gravity (g) that is ;

  • F = mg

and, the change in the position of the book that is Δx (Height)

→ Δx = Final position - Initial position

which is only the height, then the amount of work done will be calculated by :

W= mgh

  • Where W = Work Done

m = Mass of the Body

g = Acceleration due to Gravity

h = Height of Body being displaced

4 0
3 years ago
the density of water at different temperature is listed in the table above. Based on this information we can predict that the de
son4ous [18]

look it nup on google should help

5 0
3 years ago
To titrate 45.00 milliliters of an unknown HCl sample, use 25.00 milliliters of a standard 0.2000 M NaOH titrant. What is the mo
gtnhenbr [62]
V ( HCl ) = 45.00 mL in liters : 45.00 / 1000 => 0.045 L

M ( HCl ) = ?

V ( NaOH ) = 25.00 / 1000 => 0.025 L 

M ( NaOH) = 0.2000 M

number of moles NaOH :

n = M x V = 0.2000 x 0.025 => 0.005 moles of NaOH

Mole ratio:

HCl + NaOH = NaCl + H2O

1 mole HCl ---------- 1 mole NaOH
? mole HCl ---------- 0.005 moles NaOH

moles HCl = 0.005 x 1 / 1

= 0.005 moles of HCl :

M ( HCl ) = n / V

M ( HCl ) = 0.005 / 0.045

= 0.1111 M

hope this helps!




4 0
3 years ago
Calculate the pH of a solution in which one normal adult dose of aspirin (640 mg ) is dissolved in 10 ounces of water. Express y
d1i1m1o1n [39]

The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

Given data :

mass of aspirin = 640 mg = 0.640 g

volume of water = 10 ounces = 0.295735 L

molar mass of aspirin = 180.16 g/mol

moles of aspirin = mass / molar mass = 0.00355 mol

<h3>Determine the pH of the solution </h3>

First step : <u>calculate the concentration of aspirin</u>

= moles of Aspirin / volume of water

= 0.00355 / 0.295735

= 0.012 M

Given that pKa of Aspirin = 3.5

pKa = -logKa

therefore ; Ka = 10^{-3.5} = 3.162 * 10^{-4}

From the Ice table

3.162 * 10^{-4} = \frac{x + H^+}{[aspirin]}  = \frac{x^{2} }{0.012-x}

given that the value of Ka is small we will ignore -x

x² = 3.162 * 10^{-4} * 0.012

x = 1.948 * 10^{-3}  

Therefore

[ H⁺ ] = 1.948 * 10^{-3}

given that

pH = - Log [ H⁺ ]

     = - ( -3 + log 1.948 )

     = 2.71 ≈ 2.7

Hence we can conclude that The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

Learn more about Aspirin : brainly.com/question/2070753

4 0
2 years ago
Which mixture is classified as a solution?
svlad2 [7]
Salt water is considered to be a solution
5 0
3 years ago
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