Respuesta :
Acids: break down in water to shape H+ particles.
Bases: break down in water to shape OH-particles.
Lets investigate acids - A solid corrosive (HCl) versus a powerless corrosive (CH3COOH).
A solid corrosive will completely separate in water to frame H+ particles.
HCl + H2O - > H3O+ + Cl-
This response is non-reversible. After disintegration, just an exact moment grouping of HCl itself stays in the arrangement, as a large portion of the weakened HCl has broken up into particles.
Ka = [H+] [Cl-]/[HCl]
A powerless corrosive, however experiences a reversible response as it just mostly separates.
CH3COOH + H2O < - > CH3COO-+ H3O+
Ka = [CH3COO-] [H3O+]/[CH3COOH]
As Ka is a steady, as more CH3COOH is included, by Le Chatelier's Principle, in this reversible response, the framework will mean to move the balance in that capacity to invert the change. Hence, less CH3COOH than included will separate into its constituent particles.
The favourability of a response (regardless of whether forward or turn around) will depend to a great extent on the temperature and change in entropy.
The same connected for bases. The main contrast is that the base separates to frame OH-particles.
Solid and feeble bases will depend in like manner on whether the response is reversible (there are exemptions to this obviously)
NaOH + H2O - > Na+ + OH-+ H2O
Kb = [Na+] [OH-]/[NaOH]
Bases: break down in water to shape OH-particles.
Lets investigate acids - A solid corrosive (HCl) versus a powerless corrosive (CH3COOH).
A solid corrosive will completely separate in water to frame H+ particles.
HCl + H2O - > H3O+ + Cl-
This response is non-reversible. After disintegration, just an exact moment grouping of HCl itself stays in the arrangement, as a large portion of the weakened HCl has broken up into particles.
Ka = [H+] [Cl-]/[HCl]
A powerless corrosive, however experiences a reversible response as it just mostly separates.
CH3COOH + H2O < - > CH3COO-+ H3O+
Ka = [CH3COO-] [H3O+]/[CH3COOH]
As Ka is a steady, as more CH3COOH is included, by Le Chatelier's Principle, in this reversible response, the framework will mean to move the balance in that capacity to invert the change. Hence, less CH3COOH than included will separate into its constituent particles.
The favourability of a response (regardless of whether forward or turn around) will depend to a great extent on the temperature and change in entropy.
The same connected for bases. The main contrast is that the base separates to frame OH-particles.
Solid and feeble bases will depend in like manner on whether the response is reversible (there are exemptions to this obviously)
NaOH + H2O - > Na+ + OH-+ H2O
Kb = [Na+] [OH-]/[NaOH]
