A chemistry student weighs out 0.123 grams of chloroacetic acid into a 250. milliliter volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2100 moles/liter NaOH solution.
Calculate the volume of NaOH solution the student will need to add to reach the equivalence point.
See your other post above.
To calculate the volume of NaOH solution required to reach the equivalence point, we need to first determine the number of moles of chloroacetic acid in the volumetric flask.
Step 1: Calculate the number of moles of chloroacetic acid.
We are given the mass of chloroacetic acid (0.123 grams) and its molar mass can be obtained from the periodic table: Cl = 35.45 g/mol, C = 12.01 g/mol, H = 1.01g/mol, O = 16.00 g/mol.
Molar mass of chloroacetic acid = molar mass of Cl + molar mass of C + molar mass of H + molar mass of O
= (1 * 35.45) + (2 * 12.01) + (1 * 1.01) + (2 * 16.00)
= 94.41 g/mol
Number of moles of chloroacetic acid = mass of chloroacetic acid / molar mass of chloroacetic acid
= 0.123 g / 94.41 g/mol
Step 2: Calculate the volume of NaOH solution required.
We are given the concentration of NaOH solution as 0.2100 moles/liter. At the equivalence point, the moles of chloroacetic acid will react with an equal number of moles of NaOH. Therefore, the volume of NaOH solution required can be calculated using the following equation:
Volume of NaOH solution (in liters) = Number of moles of chloroacetic acid / Concentration of NaOH solution
Since we have the concentration in moles per liter, we can directly calculate the volume in liters.
Volume of NaOH solution (in liters) = Number of moles of chloroacetic acid / Concentration of NaOH solution
= 0.123 g / 94.41 g/mol / 0.2100 moles/liter
Now we have the volume in liters.
Step 3: Convert the volume to milliliters.
The volume of the volumetric flask given is in milliliters, so we need to convert the calculated volume from liters to milliliters.
Volume of NaOH solution (in milliliters) = Volume of NaOH solution (in liters) * 1000
= (0.123 g / 94.41 g/mol / 0.2100 moles/liter) * 1000
By performing the calculation, we can determine the volume of NaOH solution the student will need to add to reach the equivalence point.