A 25.0 mL aliquot of vinegar was diluted to 250 mL in a volumetric
flask . Titration of 50 mL aliquot of the diluted solution required an
average of 34.88 mLof 0.096 M NaOH. Express the acidity of vinegar
in terms of the percentage w/v of acetic acid?
CH3COOH + NaOH ==> CH3COONa + H2O
millimoles NaOH = mL x M = 34.88 mL x 0.096 M = approximately 3.35 but you should confirm that.
The equation tells you that 1 mol CH3COOH = 1 mol NaOH; therefore,
millimoles CH3COOH in the 50 mL aliquot = approx 3.35. That's the millimoles vinegar in the 50 mL; convert that to millimols acid in the 250 mL flask. That's 3.35 millimoles x (250/50) = approximately 16.8 millimoles acid in the initial 25.0 mL sample. Again, confirm that. How many grams is that. grams = mols x molar mass = 0.0168 mols x 60 g/mol = approx 1 gram acetic acid. You had that in 25.0 mL so
g/mL = 1/25 = about 0.04 g/mL or 4% w/v, You should recalculate each step. You haven't handled the number of significant figures very well; for example you give the volume to 4 places (34.88 but only give 2 for the M of NaOH as 0.096 so I've just estimated the numbers above.
Well, let's break it down! We know that a 25.0 mL aliquot of vinegar was diluted to 250 mL in a volumetric flask. So, the dilution factor is 250 mL/25.0 mL, which simplifies to 10.
Now, during titration, a 50 mL aliquot of the diluted solution required an average of 34.88 mL of 0.096 M NaOH.
To find the acidity of vinegar in terms of percentage w/v of acetic acid, we need to calculate how much acetic acid was present in the 50 mL aliquot of the diluted solution.
First, let's calculate the moles of NaOH used:
moles NaOH = concentration of NaOH (M) x volume of NaOH (L)
moles NaOH = 0.096 M x (34.88 mL / 1000 mL/L) = 0.003348 moles NaOH
Since the stoichiometry between NaOH and acetic acid is 1:1, we can say that the number of moles of acetic acid in the 50 mL aliquot is also 0.003348 moles.
To find the percentage w/v of acetic acid, we need to consider the volume of the original aliquot (25.0 mL) and the mass of acetic acid present.
The molarity of the diluted solution can be calculated:
Molarity of diluted solution = moles of acetic acid / volume of diluted solution (L)
Molarity of diluted solution = 0.003348 moles / 0.25 L = 0.013392 M
Now that we know the molarity of the diluted solution, we can calculate the mass of acetic acid in the diluted solution using the formula:
Mass of acetic acid = molarity of diluted solution x molar mass of acetic acid x volume of original aliquot (L)
Mass of acetic acid = 0.013392 M x 60.05 g/mol x (25.0 mL / 1000 mL/L) = 0.0201108 grams
Finally, to express the acidity of vinegar as a percentage w/v of acetic acid:
Percentage w/v of acetic acid = (mass of acetic acid / volume of original aliquot) x 100%
Percentage w/v of acetic acid = (0.0201108 g / 25.0 mL) x 100% = 0.0804% w/v
So, the acidity of vinegar in terms of the percentage w/v of acetic acid is approximately 0.0804%.
To find the acidity of vinegar in terms of the percentage w/v (weight/volume) of acetic acid, we need to determine the concentration of acetic acid in the diluted solution.
First, we'll calculate the number of moles of NaOH used in the titration:
Moles of NaOH = molarity of NaOH * volume of NaOH used
= 0.096 M * 34.88 mL
= 3.34848 mmol
Since the reaction between acetic acid (CH3COOH) and NaOH (sodium hydroxide) is 1:1, the number of moles of acetic acid in the diluted solution is also 3.34848 mmol.
Now we can calculate the concentration of acetic acid in the diluted solution:
Concentration of acetic acid = moles of acetic acid / volume of diluted solution
= 3.34848 mmol / 250 mL
= 0.013394 M
Next, we'll find the number of moles of acetic acid in the original 25.0 mL aliquot:
Moles of acetic acid in original solution = concentration of acetic acid * volume of original solution
= 0.013394 M * 25.0 mL
= 0.33485 mmol
Finally, we can express the acidity of vinegar in terms of the percentage w/v of acetic acid:
Percentage w/v of acetic acid = (mass of acetic acid / volume of vinegar) * 100
= (molar mass of acetic acid * moles of acetic acid / volume of vinegar) * 100
= (60.052 g/mol * 0.33485 mmol / 25.0 mL) * 100
= 0.8023%
Therefore, the acidity of the vinegar can be expressed as 0.8023% w/v of acetic acid.
To express the acidity of vinegar in terms of the percentage w/v (weight/volume) of acetic acid, we need to determine the concentration of acetic acid in the diluted solution obtained from the 25.0 mL aliquot.
Here's how you can calculate it step-by-step:
Step 1: Calculate the moles of NaOH used in the titration:
Since the concentration and volume of NaOH used are given, we can use the formula:
moles NaOH = concentration NaOH x volume NaOH
moles NaOH = 0.096 M x 34.88 mL = 3.328 moles
Step 2: Determine the moles of acetic acid in the diluted solution:
The balanced chemical equation for the reaction between acetic acid (CH3COOH) and NaOH is:
CH3COOH + NaOH → CH3COONa + H2O
From the balanced equation, we can see that the stoichiometric ratio between acetic acid and NaOH is 1:1. Therefore, the number of moles of acetic acid is also 3.328 moles.
Step 3: Calculate the concentration of acetic acid in the diluted solution:
The diluted solution's volume is 250 mL, and the concentration of acetic acid (CH3COOH) can be expressed as moles per liter (M).
moles CH3COOH = 3.328 moles
volume CH3COOH = 250 mL = 0.25 L
concentration CH3COOH = moles CH3COOH / volume CH3COOH
concentration CH3COOH = 3.328 moles / 0.25 L = 13.31 M
Step 4: Convert the concentration to percentage w/v:
The percentage w/v is the mass of solute (acetic acid) per 100 mL of solution.
mass CH3COOH = concentration CH3COOH x molar mass CH3COOH
The molar mass of acetic acid (CH3COOH) is 60.052 g/mol.
mass CH3COOH = 13.31 M x 60.052 g/mol = 799.6 g
percentage w/v CH3COOH = (mass CH3COOH / volume solution) x 100
volume solution = 250 mL + 25 mL (original aliquot) = 275 mL = 0.275 L
percentage w/v CH3COOH = (799.6 g / 0.275 L) x 100 = 290.6%
Therefore, the acidity of vinegar, expressed as the percentage w/v of acetic acid, is 290.6%.