Write the equilibrium expression, Kc , for the following chemical reaction.?

Question

Fe3+(aq) + 3OH-(aq) <==> Fe(OH)3(s)

thz

Answer 1

n general the equilibrium constant is defined in terms of activities. It is given by the product oft the product activities raised to the power their stoichiometric coefficient, divided by the product of the reactants raised to the power of their stoichiometric coefficient.
For this reaction:
K = a(Fe(OH)₃) / [ a(Fe³⁺) · (a(OH⁻))³ ]

For a component which occurs only in a pure phase like NH₄HS as solid the activity equals unity:
a(Fe(OH)₃)= 1
→
K = 1 / [ a(Fe³⁺) · (a(OH⁻))³ ]

Because for dilute aqueous solutions the activity of a solute is proportional to its concentrations for dilute aqueous solutions:
a(X) = c(X) / c_t (c_t is the total molar concentration)
you can express the equilibrium in terms of concentrations:

Kc = 1 / [ c(Fe³⁺) · (c(OH⁻))³ ]

Answer 2

you probably mean the solubility constant. The reaction is pretty much irreversible so you can't talk about equilibrium here. Normally, the equilibrium constant for the reaction you wrote would be:
Keq = [Fe(OH)3]/[Fe3+]*[OH-]^3, but since the numerator is a solidthe equation is: Keq=1/[Fe3+]*[OH-]^3, or the reverse if you write the reaction in the opposite direction.
Solubility constant on the other hand would be Ksp=[Fe3+]*[OH-]^3, which describes the equilibrium in solution between the ions and the insoluble salt.

Answer 3

For a reaction:

a*A + b*B = c*C + d*D,

the equilibrium constant can be written as:

Kc = [C]^c * [D]^d / ([A]^a * [B]^b)

where the brackets indicate the molar concentration of the species inside.

By convention, the concentration of pure solids (ones that do not vary in composition) is taken to be 1. Similarly, whenever liquid water appears in a reaction it's concentration is conventionally taken as equal to 1.

In your case, the equilibrium constant would be written as:

Kc = 1/([Fe3+]*[OH-]^3)

Answer 4

Kc = 1/[Fe3+][OH-]^3