A 10.00 ml sample of vinegar, density 1.01g/ml, was diluted to 100.0ml volume. It was found that 25.0 ml of the diluted vinegar required 24.15ml of 0.0976 M NaOH to neutralize it. Calculate the strength of CH3COOH in terms of:

Question

A 10.00 ml sample of vinegar, density 1.01g/ml, was diluted to 100.0ml volume. It was found that 25.0 ml of the diluted vinegar required 24.15ml of 0.0976 M NaOH to neutralize it. Calculate the strength of CH3COOH in terms of:

a) molarity
b) grams CH3COOH per litre
c) % Ch3COOH

For part a I used C1V1=C2V2
and found the molarity to be 0.0943M. I don't know if this is correct or not and if it isn't how do I go about doing this question?
I also don't understand how to go about completing parts b and c. I don't want the answer just the explanation on how to do the question.

Thank you for helping!!!

Answer 1

wouldn't the original vinegar be ten times that calculation, or .943M

grams per liter:
.943=grams/molmass
grams=.943*molmass

percent?
percent=massVinegar/masssolution
= massabove/liter*1liter/1.01*1000

in the last, instead of 10ml, I assumed for calculation 1 liter of the vinegar to determine percent.

Answer 2

To calculate the strength of CH3COOH in terms of molarity (part a), you correctly used the formula C1V1 = C2V2.

Given:
C1 = unknown molarity of CH3COOH
V1 = 10.00 mL (original volume of vinegar)
C2 = 0.0976 M (molarity of NaOH solution)
V2 = 24.15 mL (volume of NaOH solution required to neutralize the diluted vinegar)

Substituting the values into the equation:
C1 * 10.00 mL = 0.0976 M * 24.15 mL

Now, solve for C1:
C1 = (0.0976 M * 24.15 mL) / 10.00 mL
= 0.235296 M

Therefore, the molarity of the CH3COOH in the original vinegar solution is 0.235296 M (or approximately 0.235 M).

For part b, to calculate the grams of CH3COOH per liter, you need to convert the molarity of CH3COOH to grams of CH3COOH. The molar mass (MM) of CH3COOH is 60.05 g/mol.

To convert moles to grams, use the formula:
Grams = Moles * MM

Given:
Molarity of CH3COOH = 0.235296 M

Now, calculate the moles of CH3COOH:
Moles = Molarity * Volume
= 0.235296 mol/L * 1 L
= 0.235296 mol

Finally, calculate the grams of CH3COOH:
Grams = Moles * MM
= 0.235296 mol * 60.05 g/mol
= 14.130g

Therefore, the grams of CH3COOH per liter is 14.130 g/L.

For part c, to calculate the percentage of CH3COOH, you need to consider the mass of CH3COOH compared to the total mass of the solution.

Given:
Mass of CH3COOH = 14.130 g
Volume of vinegar = 10.00 mL (original volume)
Volume of diluted vinegar = 25.00 mL (final volume)
Density of vinegar = 1.01 g/mL

First, calculate the mass of the diluted vinegar:
Mass of diluted vinegar = Density * Volume
= 1.01 g/mL * 25.00 mL
= 25.25 g

Next, calculate the percentage of CH3COOH in the diluted vinegar:
Percentage of CH3COOH = (Mass of CH3COOH / Mass of diluted vinegar) * 100
= (14.130 g / 25.25 g) * 100
= 55.92%

Therefore, the percentage of CH3COOH in the diluted vinegar is 55.92%.