You were given 25.00 ml of an acetic acid solution of unknown concentration. You find that it requires 29.60 ml of a 0.1050 M NaOH solution to exaclty neutralize this sample.
A) What is the molarity of the acetic acid solution?
B) what is the percentage of acetic acid in the solution? Assume the density of the solution is 1 g/ml.
To solve this problem, we can use the equation for neutralization reactions:
moles of acid = moles of base
The balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:
CH3COOH + NaOH -> CH3COONa + H2O
From the equation, we can see that one mole of acetic acid reacts with one mole of sodium hydroxide.
A) To find the molarity of the acetic acid solution, we need to calculate the moles of acetic acid and the volume of the solution.
1. Calculate the moles of NaOH used:
moles of NaOH = molarity of NaOH * volume of NaOH solution
moles of NaOH = 0.1050 M * 29.60 ml
Convert the volume to liters:
29.60 ml = 29.60 ml * (1 L / 1000 ml) = 0.02960 L
moles of NaOH = 0.1050 M * 0.02960 L = 0.003108 moles
2. Since the equation is balanced with a 1:1 ratio between acetic acid and NaOH, the moles of acetic acid will be the same as the moles of NaOH used.
moles of acetic acid = 0.003108 moles
3. Calculate the molarity of the acetic acid solution:
Molarity of acetic acid = moles of acetic acid / volume of acetic acid solution
Given volume of acetic acid solution = 25.00 ml = 0.02500 L
Molarity of acetic acid = 0.003108 moles / 0.02500 L ≈ 0.1243 M
Therefore, the molarity of the acetic acid solution is approximately 0.1243 M.
B) To find the percentage of acetic acid in the solution, we can use the molarity found in part A and the molar mass of acetic acid (CH3COOH).
The molar mass of acetic acid = (12.01 g/mol) + (1.01 g/mol)x3 + (16.00 g/mol)x2
= 60.05 g/mol
1. Calculate the moles of acetic acid in the solution:
moles of acetic acid = Molarity of acetic acid * volume of acetic acid solution
moles of acetic acid = 0.1243 M * 0.02500 L = 0.003108 moles
2. Calculate the mass of acetic acid in the solution:
mass of acetic acid = moles of acetic acid * molar mass of acetic acid
mass of acetic acid = 0.003108 moles * 60.05 g/mol ≈ 0.1865 g
3. Calculate the percentage of acetic acid:
percentage of acetic acid = (mass of acetic acid / mass of solution) * 100
Given the density of the solution is 1 g/mL, the mass of the solution is equal to the volume:
mass of solution = volume of acetic acid solution = 25.00 g
percentage of acetic acid = (0.1865 g / 25.00 g) * 100 ≈ 0.746%
Therefore, the percentage of acetic acid in the solution is approximately 0.746%.
To find the molarity of the acetic acid solution, we can use the equation for neutralization:
M1V1 = M2V2
where M1 is the molarity of the acetic acid solution, V1 is the volume of the acetic acid solution, M2 is the molarity of the NaOH solution, and V2 is the volume of the NaOH solution.
First, let's convert the volumes of the solutions to liters:
V1 = 25.00 ml = 0.02500 L
V2 = 29.60 ml = 0.02960 L
Now, we can substitute the values into the equation and solve for M1:
M1(0.02500 L) = (0.1050 M)(0.02960 L)
M1 = (0.1050 M)(0.02960 L) / (0.02500 L)
M1 = 0.12384 M
Therefore, the molarity of the acetic acid solution is 0.12384 M.
To find the percentage of acetic acid in the solution, we need to determine the mass of acetic acid present. We can use the formula:
Mass = Volume x Density
Given that the density of the solution is 1 g/ml and the volume of the solution is 25.00 ml, the mass of the solution is:
Mass = 25.00 ml x 1 g/ml = 25.00 g
Next, we need to calculate the molar mass of acetic acid (CH3COOH):
C = 12.01 g/mol
H = 1.008 g/mol (3 hydrogen atoms)
O = 16.00 g/mol
Molar mass of acetic acid = (12.01 g/mol) + (1.008 g/mol x 3) + (16.00 g/mol) = 60.05 g/mol
Now, we can calculate the mass of acetic acid in the solution using the molarity and volume of the acetic acid solution:
Mass of acetic acid = Molarity x Volume x Molar mass
= 0.12384 M x 0.02500 L x 60.05 g/mol
Finally, we can calculate the percentage of acetic acid in the solution:
Percentage of acetic acid = (Mass of acetic acid / Mass of solution) x 100
= (0.12384 M x 0.02500 L x 60.05 g/mol / 25.00 g) x 100
Therefore, the percentage of acetic acid in the solution can be calculated using the given information.
HAc + NaOH ==> NaAc + H2O so it is a 1:1 titration mixture. Therefore,
mLacid x M acid = mL base x M base.
Substitute and solve for the one unknown.
B) Determine moles acetic acid, then grams.
moles = M x L
grams = moles x molar mass
%acetic acid = (grams acetic acid/mass sample)*100 = ??