What is the equilibrium concentration of chloroacetic acid in a solution prepared by dissolving 0.0314 mil of chloroacetic acid in 1.10L of water? For chloroacetic acid, Ka= 1.4X10^-3
See your other post above. Set up ICE chart and solve.
To find the equilibrium concentration of chloroacetic acid in the solution, we need to calculate the concentration of its dissociated form, which is the chloroacetate ion (C2H2ClO2-). This can be done using the equilibrium expression for the dissociation of chloroacetic acid:
Ka = [C2H2ClO2-] / [CH2ClCOOH]
First, we need to convert the mass of chloroacetic acid (0.0314 mil) to moles. The molar mass of chloroacetic acid can be calculated by summing the molar masses of its constituent elements: C (12.01 g/mol), H (1.01 g/mol), Cl (35.45 g/mol), and O (16.00 g/mol).
Molar mass of chloroacetic acid = (2 * 12.01 g/mol) + (2 * 1.01 g/mol) + 35.45 g/mol + 16.00 g/mol = 94.47 g/mol
Now, we can calculate the number of moles of chloroacetic acid:
moles of chloroacetic acid = mass / molar mass = 0.0314 mil / 94.47 g/mol
Next, we need to convert the volume of water (1.10 L) to liters:
volume in liters = 1.10 L
Using the given Ka value, and assuming the dissociation of chloroacetic acid is small compared to its initial concentration, we can set up the equilibrium expression:
1.4X10^-3 = [C2H2ClO2-] / [CH2ClCOOH]
Since the initial concentration of chloroacetic acid is proportional to its moles and volume, we can express it as:
[CH2ClCOOH] = (moles of chloroacetic acid) / (volume in liters)
Substituting the values, we get:
[CH2ClCOOH] = (0.0314 mil / 94.47 g/mol) / 1.10 L
Simplifying this expression will give us the concentration of chloroacetic acid in moles per liter.