a 10 ml vinegar sample was completely neutralized by 22.5 0.2M NaOH solution . calculate the molarity and percent acetic acid in vinegar I keep getting 2.7% acetic acid is this correct? my friend got 27% and if I am wrong where do you think I messed up?
I don't believe you did.
Is that 22.5 mL of 0.2M NaOH?
mols NaOH = M x L = 0.0225 x 0.20 = 0.00450 mols.
mols acetic acid (HAc)in the vinegar = mols NaOH (Look at the coefficients in the balanced equation.)
grams HAc = mols x molar mass = 0.00450 x 60 = about 0.270g
%HAc = g HAc/100 mL
%HAc = 0.270/10 mL = 2.70 g/100 mL = 2.70%
M = mols/L soln. Can you handle that?
Well, it seems like you and your friend are having a bit of a pickle with your calculations! But don't worry, I'm here to help. Let's break it down:
To find the molarity and percent acetic acid in vinegar, we need to consider the neutralization reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH -> CH3COONa + H2O
From the balanced equation, we can see that the stoichiometric ratio between CH3COOH and NaOH is 1:1. This means that it takes an equal amount of moles of NaOH to neutralize moles of CH3COOH.
Given that 22.5 ml of 0.2M NaOH solution neutralizes 10 ml of vinegar, we can calculate the moles of NaOH used:
0.0225 L (22.5 ml) x 0.2 mol/L = 0.0045 mol NaOH
Since the stoichiometric ratio is 1:1, this means that the moles of acetic acid (CH3COOH) in the vinegar sample is also 0.0045 mol.
Now, let's find the molarity of acetic acid in vinegar:
Molarity = moles/volume
Molarity = 0.0045 mol / 0.01 L (10 ml) = 0.45 M
Therefore, the molarity of acetic acid in the vinegar sample is 0.45 M.
To calculate the percent acetic acid, we need to consider the molecular weight of acetic acid (CH3COOH), which is approximately 60 g/mol.
Percent acetic acid = (mass of acetic acid/mass of vinegar) x 100
Percent acetic acid = (0.0045 mol x 60 g/mol) / (10 ml x density of vinegar) x 100
Here's where things might have gone awry - the density of the vinegar sample should be taken into account. So, without knowing the density, it's difficult to calculate the percent acetic acid accurately.
But if we assume the density of vinegar to be 1 g/ml, we can estimate:
Percent acetic acid = (0.0045 mol x 60 g/mol) / (10 ml x 1 g/ml) x 100
Percent acetic acid = (0.27 g / 10 g) x 100 = 2.7%
So, based on this rough estimation, your initial calculation of 2.7% acetic acid seems correct. However, keep in mind that the uncertainty in the density of vinegar may affect the accuracy of this result.
I hope this helps clarify things! Remember, chemistry can be a bit tricky, but don't let it sour your day!
To solve this problem, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH → CH3COONa + H2O
From the equation, we can see that the stoichiometric ratio between acetic acid and sodium hydroxide is 1:1.
First, let's calculate the moles of NaOH used to neutralize the vinegar sample:
10 mL of vinegar = 0.01 L of vinegar
Moles of NaOH = Molarity × Volume
Moles of NaOH = 0.2 mol/L × 0.01 L
Moles of NaOH = 0.002 mol
Since the stoichiometric ratio is 1:1, the moles of acetic acid in the vinegar sample will also be 0.002 mol.
Now, let's calculate the molarity and percent of acetic acid in the vinegar sample:
Molarity of acetic acid = Moles of acetic acid / Volume of vinegar sample
Molarity of acetic acid = 0.002 mol / 0.01 L
Molarity of acetic acid = 0.2 M
Percent acetic acid = (Moles of acetic acid / Moles of vinegar sample) × 100%
Percent acetic acid = (0.002 mol / 0.002 mol) × 100%
Percent acetic acid = 100%
Therefore, the correct answer is the percent acetic acid in vinegar is 100%. It seems like you may have made an error in your calculation. The correct answer is not 2.7% or 27%, but 100%.
To calculate the molarity and percent acetic acid in vinegar, we need to use the concept of stoichiometry.
First, let's calculate the molarity of the acetic acid (CH3COOH) in the vinegar.
The balanced chemical equation for the neutralization reaction between acetic acid and sodium hydroxide (NaOH) is:
CH3COOH + NaOH → CH3COONa + H2O
From the balanced equation, we can see that one mole of acetic acid reacts with one mole of sodium hydroxide. Therefore, the moles of sodium hydroxide used in the reaction will be equal to the moles of acetic acid present in the vinegar sample.
Given that 22.5 mL of 0.2M NaOH solution is used to neutralize the vinegar, we can calculate the moles of NaOH used:
Moles of NaOH = Molarity * Volume (in liters)
= 0.2 M * 0.0225 L
= 0.0045 moles
Since the reaction ratio is 1:1, the moles of acetic acid in the vinegar sample are also 0.0045 moles.
Next, let's calculate the molarity of acetic acid in vinegar:
Molarity = Moles of solute / Volume of solution (in liters)
= 0.0045 moles / 0.01 L
= 0.45 M
So, the molarity of acetic acid in the vinegar is 0.45M.
Now let's calculate the percent acetic acid in the vinegar:
Percent acetic acid = (moles of acetic acid / moles of solution) * 100
Since we have already determined that the moles of acetic acid are 0.0045 moles, and the volume of the vinegar sample is 0.01 L (or 10 mL), we can calculate the percent acetic acid:
Percent acetic acid = (0.0045 moles / 0.01 moles) * 100
= 45%
According to the calculations above, the percent acetic acid in the vinegar is 45%, not 2.7% or 27%.
Therefore, based on your calculations, it seems that you made an error in dividing the moles of acetic acid by the total moles of solution when calculating the percent acetic acid. Make sure to divide the moles of acetic acid by the moles of the entire solution (including both the acetic acid and the NaOH used for neutralization).