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

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.

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

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