Question

Consider the equilibrium of Al(OH)3 (s) in water. Al(OH)3 (s) = A13+ (aq) + 30H" (aq) How is the solubility of Al(OH)3 affected by adding HCl to the solution? Does the solid become more or less soluble? Explain using equilibrium and the Ksp.

Answer 1

Ksp = (Al+3)(OH-)^3

Second Answer:

Ksp = (PB2+)(Br-)^2

Answer 2

Solid become more soluble

Step-by-step explanation:

The given reaction:

`Al(OH)_3_((s))\rightleftharpoons Al^(3+)_((aq))+3OH^(-)_((aq))`

On the addition of HCl to the equilibrium, the equilibrium will shift towards right which means more dissociation of aluminum oxide. This is because the furnished hydroxide ions from aluminum oxide on dissolution now binds with the protons furnished by the acid to form water and thus there become less concentration of the hydroxide ions on the right side and to nullify this effect, the equilibrium will shift in right direction.

`OH^(-)_((aq))+H^(+)_((aq))\rightarrow H_2O_((l))`

Solubility product is defined as the product of the concentration of the ions of the solid. More the solubility product more is the solubility.

Thus, on addition of the acid, the solid become more soluble.