Pka of Ch3CO2H is 4.75
What is the equilibrium constant of the equation: Ch3Co2 (aq) + H20 (l) --> CH3CO2H(aq) + OH-(aq)
pKa = -log Ka or Ka = 10^-pKa.
Solve for Ka.
To find the equilibrium constant, we can use the equation:
K = [CH3CO2H] * [OH-] / [CH3CO2] * [H2O]
In this equation, [CH3CO2H] and [OH-] represent the molar concentrations of CH3CO2H and OH- ions, respectively, at equilibrium. [CH3CO2] represents the molar concentration of CH3CO2 at equilibrium, and [H2O] represents the molar concentration of H2O, which is considered constant (the concentration of water doesn't affect the equilibrium constant in this case).
Since the given reaction involves a weak acid, CH3CO2H, and its dissociation into ions, the equilibrium constant can be expressed in terms of the acid dissociation constant (Ka) for the weak acid:
Ka = [CH3CO2H] * [OH-] / [CH3CO2]
Given that pKa = -log10(Ka), we can calculate Ka:
Ka = 10^(-pKa) = 10^(-4.75)
Now, we have Ka, which represents the equilibrium constant of the dissociation of CH3CO2H:
Ka = [CH3CO2] * [OH-] / [CH3CO2H]
To find the equilibrium constant for the given reaction:
K = 1 / Ka = 1 / (10^(-4.75))
Using a calculator, we can solve this equation to find the equilibrium constant, K.