Calculate the molar concentration of acetic acid (CH3COOH) in a 5.00-mL sample of vinegar (density is 1.00 g/mL) if it is titrated with 25.00 mL of NaOH.
Determine moles acetic acid. moles NaOH = M NaOH x L NaOH.
Moles acetic acid = moles NaOH.
M = # moles/L of solution.
To calculate the molar concentration of acetic acid in the vinegar sample, you will need to determine the number of moles of acetic acid present in the 5.00 mL sample and then divide it by the volume of the sample.
Here are the steps to get the answer:
Step 1: Determine the number of moles of NaOH used in the titration.
To do this, you will need to know the concentration of the NaOH solution. Assuming the concentration of NaOH is given in moles per liter (M), you can use the formula:
moles of NaOH = concentration of NaOH (M) × volume of NaOH (L)
In this case, the volume of NaOH used is 25.00 mL, which is equal to 0.025 L.
Step 2: Calculate the number of moles of acetic acid.
From the balanced chemical equation:
CH3COOH + NaOH → CH3COONa + H2O
We can see that one mole of acetic acid reacts with one mole of NaOH. Therefore, the number of moles of acetic acid is equal to the number of moles of NaOH used in the titration.
Step 3: Calculate the molar concentration of acetic acid.
To determine the molar concentration, divide the number of moles of acetic acid by the volume of the vinegar sample.
molar concentration of acetic acid = moles of acetic acid / volume of vinegar sample (in L)
The sample volume is given as 5.00 mL, which is equal to 0.005 L.
Now, you can substitute the calculated values into the equation to find the molar concentration of acetic acid.
Thank you very much.
The molarity of the NaOH solution is 0.162.
So I did the following:
25.00 mL NaOH*(1 L / 1000 mL)*(0.160 mol NaOH / 1 L NaOH)*(1 mol CH3COOH / 1 mol NaOH) = 0.004 mol CH3COOH
(0.004 mol CH3COOH / 5.00 mL) * (1000 mL / 1 L) = 0.8 M CH3COOH
Now I'm supposed to determine the mass percent of the acetic acid in the vinegar sample. I assume this is why I was given the density...
how do I use the concentration of acetic acid and the density of vinegar to determine the mass percent?
typo...
"The molarity of the NaOH solution is 0.160." The number in the calculations is correct.
moles CH3COOH x molar mass acetic acid = grams CH3COOH.
(g CH3COOH/100 g soln)*100 = mass percent. which you can do one of two ways.
I have 0.24 g/5 g so how much is in 100?
That will be 0.24 x 100/5 = 0.24 x 20 = 4.8 which is 4.8%.
The other way, I think, is simpler:
(0.24/5)*100 =