The titration of 10ml sample of vinegar requires 30ml of 0.20M of NaOH solution. Find molarity and mass percent concentration of the acetic acid.
NaOH + CH3COOH ==> H2O + CH3COONa
moles NaOH = M x L = 0.20 x 0.030 L = 0.006
You see 1 mol CH3COOH to 1 mol NaOH; therefore, 0.006 mols NaOH is due to 0.006 mols CH3COOH (vinegar--;i.e., acetic acid).
M acetic acid = moles/L = 0.006/0.010 = 0.60 M.
To do mass percent you need the mass of the 10 mL (you don't have that and no way to get it that I see). You also need the mass of the acetic acid in the sample. That is grams = mols x molar mass = ?
Oh, vinegar and titration, sounds like chemistry fun! Alright, let's dive into it.
To find the molarity of acetic acid (CH3COOH) in vinegar, we need to use the equation:
Molarity of acid × Volume of acid = Molarity of base × Volume of base
Given that the volume of the vinegar (acetic acid) is 10 ml and the volume of the NaOH solution (base) is 30 ml, and the molarity of the NaOH solution is 0.20 M, we can rearrange the equation to solve for the molarity of acetic acid:
Molarity of acid = (Molarity of base × Volume of base) / Volume of acid
Molarity of acid = (0.20 M × 30 ml) / 10 ml
Molarity of acid = 0.60 M
So, the molarity of acetic acid in the vinegar is 0.60 M.
Now, let's move on to the mass percent concentration of acetic acid in vinegar. The formula to calculate mass percent concentration is:
Mass percent concentration = (Mass of solute / Mass of solution) × 100
To find the mass of acetic acid (solute) in 10 ml of vinegar (solution), we need to know the density of vinegar. Since vinegar density can vary, let's assume it's approximately 1 g/ml for simplicity.
Mass of acetic acid = Volume of vinegar × Density of vinegar
Mass of acetic acid = 10 ml × 1 g/ml
Mass of acetic acid = 10 g
To find the mass percent concentration, we divide the mass of acetic acid by the mass of the solution (which is the mass of acetic acid + any additional solvents):
Mass percent concentration = (10 g / 10 g) × 100
Mass percent concentration = 100%
So, the molarity of acetic acid is 0.60 M, and the mass percent concentration of acetic acid is 100%. Time to grab some popcorn and celebrate the chemistry of vinegar! 🎉
To find the molarity and mass percent concentration of the acetic acid in the vinegar sample, we need to use the balanced chemical equation and the volume and molarity of the NaOH solution used.
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 the ratio between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is 1:1.
Given:
Volume of NaOH solution (V1) = 30 mL = 0.03 L
Molarity of NaOH solution (M1) = 0.20 M
Volume of vinegar sample (V2) = 10 mL = 0.01 L
To find the moles of NaOH used, we can use the following equation:
moles of NaOH (n1) = Molarity x Volume
= M1 x V1
= 0.20 M x 0.03 L
= 0.006 moles
Since the ratio between acetic acid and NaOH is 1:1, the moles of acetic acid (CH3COOH) in the vinegar sample (n2) is also 0.006 moles.
To find the molarity (M2) of the acetic acid in the vinegar sample, we can use the formula:
Molarity (M2) = moles (n2) / Volume (V2)
= 0.006 moles / 0.01 L
= 0.60 M
Therefore, the molarity of the acetic acid in the vinegar sample is 0.60 M.
To find the mass percent concentration of the acetic acid, we need to know the molar mass of acetic acid, which is calculated as follows:
Molar mass of acetic acid = (12.01 g/mol x 2) + (1.01 g/mol x 4) + (16.00 g/mol + 1.01 g/mol)
= 60.05 g/mol
Mass of acetic acid (m2) = moles (n2) x Molar mass
= 0.006 moles x 60.05 g/mol
= 0.3603 g
Mass percent concentration = (mass of solute / mass of solution) x 100%
= (0.3603 g / 10 g) x 100%
= 3.60%
Therefore, the mass percent concentration of acetic acid in the vinegar sample is 3.60%.
To find the molarity and mass percent concentration of acetic acid in the vinegar sample, we will use the concept of stoichiometry and the equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) as follows:
CH3COOH + NaOH -> CH3COONa + H2O
First, let's determine the number of moles of NaOH used in the titration:
Moles of NaOH = Molarity of NaOH x Volume of NaOH solution
= 0.20 M x 30 ml
= 6.00 mmol
According to the balanced equation, the stoichiometric ratio between NaOH and acetic acid is 1:1. Therefore, the number of moles of acetic acid present in the vinegar sample is also 6.00 mmol.
Next, we need to determine the mass of acetic acid in the vinegar sample. The molar mass of acetic acid (CH3COOH) is 60.05 g/mol.
Mass of acetic acid = Moles of acetic acid x Molar mass of acetic acid
= 6.00 mmol x 60.05 g/mol
= 0.360 g
To find the molarity of acetic acid in the vinegar sample, we need to consider the volume of the vinegar sample used in the titration. In this case, the volume of the vinegar sample is 10 ml.
Molarity of acetic acid = Moles of acetic acid / Volume of vinegar sample
= 6.00 mmol / 10 ml
= 0.600 M
Finally, to find the mass percent concentration of acetic acid in the vinegar sample, we use the formula:
Mass percent concentration = (Mass of acetic acid / Mass of vinegar sample) x 100
Since we have already found the mass of acetic acid as 0.360 g, we need to determine the mass of the vinegar sample. Assuming the density of vinegar is 1 g/ml, the mass of the vinegar sample can be calculated:
Mass of vinegar sample = Volume of vinegar sample x Density of vinegar
= 10 ml x 1 g/ml
= 10 g
Now we can calculate the mass percent concentration:
Mass percent concentration = (0.360 g / 10 g) x 100
= 3.60%
Therefore, the molarity of acetic acid in the vinegar sample is 0.600 M, and the mass percent concentration of acetic acid is 3.60%.