chemistry student weighs out 0.123 grams of chloroacetic acid (HCH2ClCO2) into a 250. milliliter volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1200 moles/liter NaOH solution.
Calculate the volume of NaOH solution the student will need to add to reach the equivalence point.
moles chloroacetic acid = M x L
moles acid = moles NaOH
M NaOH = moles NaOH/L NaOH. You have M NaOH and M NaOH, solve for L NaOH and convert to mL if you wish.
To calculate the volume of NaOH solution needed to reach the equivalence point, we first need to determine the number of moles of chloroacetic acid (HCH2ClCO2) in 0.123 grams.
The molar mass of chloroacetic acid can be calculated by adding up the atomic masses of its constituent atoms:
H (1.0079 g/mol) + C (12.01 g/mol) + H (1.0079 g/mol) + Cl (35.45 g/mol) + C (12.01 g/mol) + O (16.00 g/mol) + O (16.00 g/mol) = 78.54 g/mol
Using the molar mass, we can determine the number of moles of chloroacetic acid by dividing the mass by the molar mass:
0.123 g / 78.54 g/mol ≈ 0.001567 mol
Since the molar ratio between chloroacetic acid and sodium hydroxide (NaOH) is 1:1 in the balanced chemical equation, the number of moles of NaOH required to neutralize the chloroacetic acid is also 0.001567 mol.
Now, we can calculate the volume of 0.1200 moles/liter (M) NaOH solution needed to provide 0.001567 moles of NaOH:
Volume (L) = moles / Molarity
Volume (L) = 0.001567 mol / 0.1200 mol/L
Volume (L) ≈ 0.01306 L
Finally, we convert the volume from liters to milliliters:
Volume (mL) = 0.01306 L * 1000 mL/L
Volume (mL) ≈ 13.06 mL
Therefore, the student will need to add approximately 13.06 milliliters of NaOH solution to reach the equivalence point.