A 10.0 mL of vinegar, an aqueous solution of acetic acid (HC2H3O2), is titrated with .5062M NaOH, and 16.58mL is required to reach the equivalence point.
A. What is the molarity of the acetic acid?
B. If the density of the vinegar is 1.006 g/cm^3, what is the mass percent of acetic acid in the vinegar?
Acetic acid is CH3COOH (HC2H3O2). Its the right end H that is acidic.
CH3COOH + NaOH ==> CH3COONa + HOH
mols NaOH required = M x L = ??
mols acetic acid must be the same since the reaction is 1 mol CH3COOH to 1 mol NaOH.
Molarity CH3OOH = # mols/L = mols/0.10 = xx.
For percent, convert mols CH3COOH from above to grams. Convert 10.0 mL to grams using the density to give the sample mass.
percent CH3COOH = [grams CH3COOH/mass sample]*100 = ??
Post your work if you get stuck.
Ok, so far this is what i've done.
A. to get moles of NaOH, i multiplied 0.5062M x 0.01658L=8.39x10^-3, then i divided this number by .010L, i got .8392M
B. (8.392x10^-3)x(60.052g HC2H3O2)=.5039g, after this part i got stuck.
You're almost home. You have grams and that is correct except I would have carried it out to another place to get 0.50399 which rounds to 0.5040 g. (The molarity I calculated as 0.83928 which rounds to 0.8393 but I didn't round and re-enter numbers in my calculator.
%acetic acid = [grams/mass sample]*100
You have grams from above.
To obtain mass of the sample,
mass = volume x density
10 cc x 1.006 g/cc = ??
To solve this problem, we will use the concept of stoichiometry and the balanced chemical equation of the reaction between acetic acid (HC2H3O2) and sodium hydroxide (NaOH). The balanced equation is:
HC2H3O2 + NaOH -> NaC2H3O2 + H2O
A. To find the molarity of acetic acid, we need to determine the number of moles of NaOH used in the titration. We can calculate this using the equation:
Moles of NaOH = Molarity of NaOH x Volume of NaOH
Moles of NaOH = 0.5062 M x 16.58 mL = 0.5062 M x 0.01658 L = 0.008390 moles
Since the stoichiometry of the reaction is 1:1 between NaOH and acetic acid, the number of moles of acetic acid used in the reaction is also 0.008390 moles.
Now, we can calculate the molarity of acetic acid:
Molarity of acetic acid = Moles of acetic acid / Volume of acetic acid
The volume of acetic acid is given as 10.0 mL, which we need to convert to liters:
Volume of acetic acid = 10.0 mL = 0.0100 L
Molarity of acetic acid = 0.008390 moles / 0.0100 L = 0.839 M
Therefore, the molarity of acetic acid is 0.839 M.
B. To calculate the mass percent of acetic acid in the vinegar, we need to determine the mass of acetic acid present in the given volume of vinegar and then express it as a percentage of the total mass of the vinegar.
First, we need to calculate the moles of acetic acid in the given volume of vinegar (10.0 mL). We can use the molarity and volume of acetic acid calculated in part A to find the moles:
Moles of acetic acid = Molarity of acetic acid x Volume of acetic acid
Moles of acetic acid = 0.839 M x 0.0100 L = 0.00839 moles
Next, we need to convert the moles of acetic acid to grams using its molar mass. The molar mass of acetic acid (HC2H3O2) is:
Molar mass of HC2H3O2 = (2 x atomic mass of carbon) + (4 x atomic mass of hydrogen) + (2 x atomic mass of oxygen)
= (2 x 12.01 g/mol) + (4 x 1.01 g/mol) + (2 x 16.00 g/mol)
= 60.05 g/mol
Mass of acetic acid = Moles of acetic acid x Molar mass of acetic acid
= 0.00839 moles x 60.05 g/mol
= 0.504 g
Finally, we can calculate the mass percent of acetic acid in the vinegar:
Mass percent = (Mass of acetic acid / Mass of vinegar) x 100
The density of vinegar is given as 1.006 g/cm^3, and the volume of the vinegar is 10.0 mL. We can calculate the mass of the vinegar using the density:
Mass of vinegar = Volume of vinegar x Density of vinegar
= 10.0 mL x 1.006 g/cm^3
= 10.06 g
Mass percent = (0.504 g / 10.06 g) x 100
= 5.01%
Therefore, the mass percent of acetic acid in the vinegar is 5.01%.