Write the equilibrium expression for the Ka of Cl3CCO2H
Cl3C-COOH --> H^+ + Cl3COO^-
Ka = (H^+)(Cl3COO^-)/(CL3C-COOH)
Trichloroacetic acid is all but a very strong acid. It isn't ionized 100%. The pKa is 0.66
The equilibrium expression for the equilibrium constant Ka of Cl3CCO2H (trichloroacetic acid) is given by:
Ka = [H+][Cl3CCO2-]/[Cl3CCO2H]
Where [H+] represents the concentration of H+ (hydronium ions), [Cl3CCO2-] represents the concentration of the conjugate base of Cl3CCO2H, and [Cl3CCO2H] represents the concentration of trichloroacetic acid itself.
To write the equilibrium expression for the Ka of Cl3CCO2H, we first need to understand what Ka represents. Ka stands for the acid dissociation constant and is used to quantify the strength of an acid in a solution.
The general form of the equilibrium expression for an acid dissociation is:
Ka = [H+][A-]/[HA]
In this expression:
- [H+] represents the concentration of hydrogen ions (protons) in the solution
- [A-] represents the concentration of the conjugate base of the acid
- [HA] represents the concentration of the acid itself
Now, let's apply this to Cl3CCO2H. The compound Cl3CCO2H is also known as trichloroacetic acid. When it dissolves in water, it donates a hydrogen ion (H+) to the solution, forming a conjugate base Cl3CCO2-. Therefore, the equilibrium expression for the Ka of Cl3CCO2H is:
Ka = [H+][Cl3CCO2-]/[Cl3CCO2H]
Please note that the concentrations mentioned in the expression can be in either molarity (M) or any other concentration units as long as they are consistent.