10 ml sample of vinegar an aqueous solution 0f acetic acid( HC2H3O2) is titrated with 0.5062 M and 16.58 ml is required to reach equivalence point what is the molarity of the acetic acid
b. if the density of vinegar is 1.006 g/cm3 what is the mass percent of acetic acid of vinegar
Step 1: Calculate the number of moles of NaOH used in the titration.
Molarity (M) = moles (n) / volume (V)
0.5062 M NaOH is used to neutralize 16.58 ml of vinegar.
moles (NaOH) = Molarity (NaOH) x volume (vinegar)
= 0.5062 mol/L x 0.01658 L
≈ 0.0084 mol NaOH
Step 2: Use the balanced chemical equation to determine the moles of acetic acid (HC2H3O2).
NaOH (aq) + HC2H3O2 (aq) → H2O (l) + NaC2H3O2 (aq)
From the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of HC2H3O2. So, the number of moles of HC2H3O2 is also 0.0084 mol.
Step 3: Calculate the molarity of the acetic acid (HC2H3O2).
Molarity (HC2H3O2) = moles (HC2H3O2) / volume (V)
volume (V) = 10 ml = 0.01 L
Molarity (HC2H3O2) = 0.0084 mol / 0.01 L
= 0.84 M
Therefore, the molarity of the acetic acid (HC2H3O2) is 0.84 M.
b. To calculate the mass percent of acetic acid in vinegar, we need to know the mass of acetic acid and the total mass of the vinegar.
Step 1: Calculate the mass of acetic acid.
molarity (HC2H3O2) = moles (HC2H3O2) / volume (V)
moles (HC2H3O2) = molarity (HC2H3O2) x volume (V)
= 0.84 mol/L x 0.010 L
= 0.0084 mol HC2H3O2
The molar mass of acetic acid (HC2H3O2) is as follows:
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
molar mass (HC2H3O2) = (2 x 1.01 g/mol) + (2 x 12.01 g/mol) + (4 x 16.00 g/mol)
= 60.05 g/mol
mass (HC2H3O2) = moles (HC2H3O2) x molar mass (HC2H3O2)
= 0.0084 mol x 60.05 g/mol
≈ 0.504 g HC2H3O2
Step 2: Calculate the total mass of vinegar.
density = mass / volume
mass = density x volume
Given that the density of vinegar is 1.006 g/cm^3 and the volume of the vinegar is 10 mL:
mass (vinegar) = 1.006 g/cm^3 x 10 mL
Since 1 mL = 1 cm^3, the mass of the vinegar is:
mass (vinegar) = 1.006 g/cm^3 x 10 cm^3
= 10.06 g
Step 3: Calculate the mass percent of acetic acid.
mass percent = (mass (HC2H3O2) / mass (vinegar)) x 100%
mass percent = (0.504 g / 10.06 g) x 100%
≈ 5.0%
Therefore, the mass percent of acetic acid in vinegar is approximately 5.0%.
To find the molarity of acetic acid in the vinegar solution, you can use the equation:
M1V1 = M2V2
Where:
M1 = molarity of the NaOH solution (0.5062 M)
V1 = volume of NaOH solution used (16.58 ml or 0.01658 L)
M2 = molarity of acetic acid
V2 = volume of acetic acid solution used (10 ml or 0.01 L)
Rearranging the equation, we have:
M2 = (M1V1) / V2
= (0.5062 M)(0.01658 L) / 0.01 L
= 0.837 M (rounded to three decimal places)
Therefore, the molarity of the acetic acid in the vinegar solution is approximately 0.837 M.
Next, to calculate the mass percent of acetic acid in vinegar, you'll need to determine the mass of acetic acid present in the 10 ml sample.
Given that the density of vinegar is 1.006 g/cm3, you can find the mass of vinegar using the formula:
Mass = Volume x Density
= 10 ml x 1.006 g/cm3 (since 1 ml = 1 cm3)
= 10.06 g
Since the density of vinegar is almost equal to the density of water (1 g/cm3), you can assume that the density of the vinegar solution is close to that of water. Hence, the mass of the vinegar solution is approximately equal to its volume.
Now, to find the mass of acetic acid, you can use the molarity and molar mass of acetic acid (CH3COOH).
Molar mass of acetic acid (CH3COOH) = (12.01 g/mol x 2) + (1.01 g/mol x 4) + 16.00 g/mol
= 60.05 g/mol
Mass of acetic acid = Molarity x Volume x Molar mass
= 0.837 M x 0.01 L x 60.05 g/mol
= 0.502 g
Finally, to calculate the mass percent of acetic acid, divide the mass of acetic acid by the mass of vinegar and multiply by 100:
Mass percent = (Mass of acetic acid / Mass of vinegar) x 100
= (0.502 g / 10.06 g) x 100
= 4.99%
Therefore, the mass percent of acetic acid in the vinegar solution is approximately 4.99%.
HAc + NaOH ==> NaAc + H2O
mols NaOH = M x L = estimated 0.008 but you need a more accurate answer for this and all of the other estimates that follow.
From the equation.
0.008 mols NaOH = 0.008 mols HAc
M = mols/L = 0.008/0.010L = about 0.8M
g HAc = mols x molar mass = approx 0.5g
mass of 10 mL = 1.006g/mL x 10 mL =?
mass% = g solute/g solution = about 5%?