2.50 mL of ethyl acetate and 2.50 mL of distilled water are added to the 5.00 mL of 3 Molar HCl and the mixture is allowed to sit for a week. At the end of this time, 28.30 mL of 1.000 Molar NaOH was needed to titrate the solution. (The HCl was the same as used in 1a. This value was 3.44 Molar HCl)
A) How many millimoles of acetic acid were present at equilibrium?
B) How many millimoles of water were present at equilibrium?
C) How many millimoles of ethyl acetate were present at equilibrium?
I have no idea where to go with this for my lab and getting help on this section would most likely help me finish the lab. So any help would be greatly appreciated!
To solve this problem, we need to use the concept of stoichiometry and the balanced chemical equation. The balanced chemical equation for the reaction between ethyl acetate (CH3COOC2H5) and hydrochloric acid (HCl) is:
CH3COOC2H5 + HCl → CH3COOH + C2H5Cl
Step 1: Calculate the moles of HCl used:
Given the concentration of HCl is 3 M and the volume used is 5.00 mL. We can use the equation:
moles of HCl = concentration of HCl × volume of HCl
= 3 mol/L × (5.00 × 10^-3 L)
= 0.015 mol
Step 2: Use stoichiometry to determine the moles of acetic acid (CH3COOH) produced:
From the balanced equation, we know that 1 mole of HCl reacts with 1 mole of acetic acid (CH3COOH). Therefore, the moles of acetic acid produced will be equal to the moles of HCl used.
moles of acetic acid = 0.015 mol
Step 3: Convert moles to millimoles:
The prefix "milli-" means one-thousandth, so one millimole is equal to 0.001 moles.
millimoles of acetic acid = 0.015 mol × 1000
= 15 millimoles
Therefore, at equilibrium, there were 15 millimoles of acetic acid present.
Step 4: Calculate the moles of NaOH used:
Given the concentration of NaOH is 1.000 M and the volume used is 28.30 mL. Using the same equation as before:
moles of NaOH = concentration of NaOH × volume of NaOH
= 1 mol/L × (28.30 × 10^-3 L)
= 0.0283 mol
Step 5: Use stoichiometry to determine the moles of ethyl acetate (CH3COOC2H5) present:
From the balanced equation, we know that 1 mole of ethyl acetate reacts with 1 mole of NaOH. Therefore, the moles of ethyl acetate present will be equal to the moles of NaOH used.
moles of ethyl acetate = 0.0283 mol
Step 6: Convert moles to millimoles:
Similar to the previous steps, we can convert moles to millimoles using the equation:
millimoles of ethyl acetate = 0.0283 mol × 1000
= 28.3 millimoles
Therefore, at equilibrium, there were 28.3 millimoles of ethyl acetate present.
Step 7: Calculate the moles of water (H2O) present:
From the balanced equation, we know that for every 1 mole of ethyl acetate that reacts, 1 mole of water is produced. Therefore, the moles of water present will be equal to the moles of ethyl acetate used.
moles of water = 0.0283 mol
Step 8: Convert moles to millimoles:
Using the same conversion as before:
millimoles of water = 0.0283 mol × 1000
= 28.3 millimoles
Therefore, at equilibrium, there were 28.3 millimoles of water present.
To summarize:
A) The number of millimoles of acetic acid present at equilibrium is 15 millimoles.
B) The number of millimoles of water present at equilibrium is 28.3 millimoles.
C) The number of millimoles of ethyl acetate present at equilibrium is also 28.3 millimoles.
To answer these questions, we need to use the principles of stoichiometry and chemical equilibrium. Let's break down each part step by step:
A) To determine the millimoles of acetic acid present at equilibrium, we need to understand the reaction that takes place between ethyl acetate and water. Ethyl acetate hydrolyzes in water to form acetic acid and ethanol. The balanced chemical equation is:
CH3COOC2H5 + H2O ⇌ CH3COOH + C2H5OH
Given that 2.50 mL of ethyl acetate is added, we need to convert this volume to millimoles. Since 1 mole of any substance contains 1000 millimoles, we can use the molar volume of ethyl acetate to calculate the millimoles:
Molar volume of ethyl acetate = 2.50 mL x (1 mole / molar volume)
Next, we need to determine the number of millimoles of acetic acid produced at equilibrium. Since the reaction is in equilibrium, we assume it goes to completion. Therefore, the number of millimoles of acetic acid produced is equal to the number of millimoles of ethyl acetate present at equilibrium.
B) To find the millimoles of water present at equilibrium, we need to consider the balanced equation again. For each mole of ethyl acetate hydrolyzed, one mole of water is consumed. Therefore, we can assume that the millimoles of water present at equilibrium are equal to the millimoles of ethyl acetate added.
C) The millimoles of ethyl acetate at equilibrium can be determined using the balanced equation. Since the reaction is in equilibrium, the millimoles of ethyl acetate remaining at equilibrium can be calculated by subtracting the millimoles of ethyl acetate that hydrolyzed (producing acetic acid and ethanol) from the initial millimoles of ethyl acetate added.
Now, to calculate the millimoles of each substance, we need to determine the concentrations using the volume of NaOH used for titration and the known molarity:
Molarity of NaOH x Volume of NaOH used (in liters) = Millimoles of substance
In this case, the molarity of NaOH is given as 1.000 M, and the volume used is 28.30 mL.
Using the above principles and calculations, you should be able to find the millimoles of acetic acid, water, and ethyl acetate present at equilibrium in the given solution.