A 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.
To calculate the volume of NaOH solution required to reach the equivalence point, we first need to determine the number of moles of chloroacetic acid present in the volumetric flask.
The molar mass of chloroacetic acid is:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16.00 g/mol
Adding the molar masses of each element in the compound, we get:
12.01 + 1.01 + 35.45 + 12.01 + 16.00 + 16.00 + 16.00 = 108.48 g/mol
Now we can calculate the number of moles of chloroacetic acid in 0.123 grams:
moles = mass / molar mass
moles = 0.123 g / 108.48 g/mol
Next, we need to determine the stoichiometric ratio between the chloroacetic acid and the NaOH solution. From the balanced chemical equation, we can see that the ratio of chloroacetic acid to NaOH is 1:1.
Now, we can calculate the volume of NaOH solution required to reach the equivalence point using the moles of chloroacetic acid and the given concentration of NaOH:
volume = moles / concentration
Given that the concentration of NaOH solution is 0.1200 moles/L, and since the desired volume is in liters, we need to convert the volume in milliliters to liters:
volume = moles / concentration
= (0.123 g / 108.48 g/mol) / (0.1200 mol/L)
= 0.001132 mol / 0.1200 mol/L
≈ 0.00943 L
Finally, to convert the volume from liters to milliliters:
volume = 0.00943 L * 1000 mL/L
= 9.43 mL
Therefore, the student will need to add approximately 9.43 mL of NaOH solution to reach the equivalence point.
To calculate the volume of NaOH solution required to reach the equivalence point, we need to use the concept of stoichiometry. The balanced equation for the reaction between chloroacetic acid (HCH2ClCO2) and NaOH is:
HCH2ClCO2 + NaOH -> H2O + NaCH2ClCO2
From the balanced equation, we can see that the molar ratio between chloroacetic acid and NaOH is 1:1. This means that for every 1 mole of chloroacetic acid, 1 mole of NaOH is required to neutralize it.
First, we need to determine the number of moles of chloroacetic acid in the 0.123 grams using its molar mass. The molar mass of chloroacetic acid (HCH2ClCO2) is:
(1 * 1.007) + (3 * 1.008) + (1 * 35.453) + (1 * 12.011) + (2 * 16.00) = 94.50 g/mol
Number of moles of chloroacetic acid = Mass / Molar mass
= 0.123 g / 94.50 g/mol = 0.0013 mol
Since the molar ratio between chloroacetic acid and NaOH is 1:1, we need 0.0013 moles of NaOH to neutralize the chloroacetic acid.
Now, we can calculate the volume of NaOH solution required using its concentration:
Concentration of NaOH solution = 0.1200 mol/L
We can use the following formula to calculate the volume (V) of a solution:
V = n / C
Where:
V = volume (in liters)
n = number of moles
C = concentration (in moles per liter)
Substituting the values into the formula, we get:
V = 0.0013 mol / 0.1200 mol/L = 0.0108 L
Finally, we need to convert the volume to milliliters (mL) since the given flask is in milliliters:
Volume of NaOH solution required = 0.0108 L * 1000 mL/L = 10.8 mL
Therefore, the student will need to add 10.8 mL of NaOH solution to reach the equivalence point.