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AnnyKZ [126]
3 years ago
13

Why amino acids are soluble in pH=3 or pH=10 solutions more than it dissolves in pH=7 solutions?

Chemistry
1 answer:
Ray Of Light [21]3 years ago
8 0

Answer:

Solubility is Affected by pH

The pH of an aqueous solution can affect the solubility of the solute. By changing the pH of the solution, you can change the charge state of the solute.

At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent. At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.

An isoelectric point is the pH at which an amino acid exists as its zwitterion. A zwitterion is the dipolar ionic form of an amino acid. ... If the pH is lower (in acidic conditions) than the isoelectric point then the amino acid acts as a base and accepts a proton at the amino group. This gives it a positive change.

An amino acid is usually more soluble in aqueous solvent at pH extremes than it is at a pH near the isolelectric point of the amino acid. (Note that this does not mean that the amino acid is insoluble at a pH near its pI.)

Which of the following statements correctly explains this phenomenon?

(Select all that apply.)

The neutral charge of an amino acid molecule at its isoelectric point will make the molecule hydrophobic.

At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent.

At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.

At pH values far from the isoelectric point, individual amino acid molecules have greater kinetic energy, thus more readily stay in solution.

<h2>Please mark me as brainliest</h2>

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What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +
nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ +  S₂O₃²⁻  + H₂O  → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻         Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

<u>5H₂O</u> + S₂O₃²⁻ → 2SO₄²⁻ + <u>10H⁺</u> + <em>8e</em>-  Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + <em>8e</em>-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻  → 2SO₄²⁻ + 10H⁺ + <em>8e</em>- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

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