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Gnesinka [82]
2 years ago
5

Hydrophobic molecules (like oil) tend to self-associate in water rather than dissolve. What is the MAJOR energetic contribution

to this property.
Chemistry
2 answers:
Inga [223]2 years ago
8 0

Answer:

The self association maximizes the entropy of the water molecules

Explanation:

Nonpolar molecules, like Oil, do not dissolve easily in water. They are described as hydrophobic, or water fearing. When put into polar environments, such as water, nonpolar molecules stick together and form a tight membrane, preventing water from surrounding the molecule.

Moreso, the hydrophobic effect was found to be entropy-driven at room temperature because of the reduced mobility of water molecules in the solvation shell of the non-polar solute.

anyanavicka [17]2 years ago
4 0

Answer: The MAJOR energetic contribution to this property is the self association that maximizes the entropy of the water molecules.

Explanation:

Hydrophobic molecules (like oil) tend to self-associate in water rather than dissolve in it. The MAJOR energetic contribution to this property is the self association of this oil which increases the degree of disorderliness of the water molecules.

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100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.
wolverine [178]

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a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>
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