Calculate the analytical and equilibrium molar concentrations of the solute species in an aqueous solution that contains 285 mg of trichloroacetic acid , CI3CCOOH ( 163.4 g / mol ) in 10 ml ( the acid is 73 % ionized in water )
analytical concentration:
mols trichloroacetic acid = grams/molar mass = 0.285 g/163.4 = 0.00174
The M = molarity = moles/L = 0.00174 moles/0.010 L of solution.
equilibrium concentrations:
.................Cl3CCOOH --> H^+ + [Cl3CCOO^-]
I.................0.00174 M...... 0..............0
C...................-x...................x...............x
E.................0.00174-x.........x...............x
The problem tells you that the acid is 73% ionized; therefore, x = 0.00174*0.73 = ? and evaluate 0.00174 - x.