Could someone help me with this question?
A vinegar sample was analyzed and found to be 5.88% acetic acid by volume. The density of pure acetic acid is 1.409g/ml. What volume of 0.200 M NaOH would be required to titrate a 2 ml aliquot of this vinegar.
5.88% v/v means 5.88 mL acetic acid/100 mL solution. m = v*d; therefore 5.88 mL x 1.409 g/mL = grams acetic acid.
Convert grams to moles.
moles = grams/molar mass.
Convert to M = moles/L of soln.
Then write the equation.
CH3COOH + NaOH ==> CH3COONa + H2O
moles CH3COOH frm above.
moles NaOH must be the same since the reaction is 1 mole CH3COOH to 1 mole NaOH.
moles CH3COOH = M x L
moles NaOH = M x L.
M x L = M x L
You have M CH3COOH and L CH3COOH and M NaOH. Solve for L NaOH.
To find the volume of 0.200 M NaOH required to titrate a 2 ml aliquot of the vinegar, we can use the concept of stoichiometry and the equation for the reaction between acetic acid and sodium hydroxide.
The balanced chemical equation for the reaction is:
CH3COOH + NaOH → CH3COONa + H2O
According to the equation, 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.
First, let's calculate the number of moles of acetic acid in the 2 ml aliquot of vinegar:
Volume of acetic acid = Volume of vinegar sample * Percentage of acetic acid / 100
Volume of acetic acid = 2 ml * 5.88 / 100 = 0.1176 ml
Since the density of pure acetic acid is 1.409 g/ml, we can calculate the mass of acetic acid:
Mass of acetic acid = Volume of acetic acid * Density of acetic acid
Mass of acetic acid = 0.1176 ml * 1.409 g/ml = 0.1656 g
Next, we can calculate the number of moles of acetic acid:
Number of moles of acetic acid = Mass of acetic acid / Molar mass of acetic acid
The molar mass of acetic acid (CH3COOH) is 60.052 g/mol:
Number of moles of acetic acid = 0.1656 g / 60.052 g/mol ≈ 0.00276 mol
Since the balanced equation tells us that the molar ratio of acetic acid to sodium hydroxide is 1:1, the number of moles of sodium hydroxide required is also 0.00276 mol.
Now, let's calculate the volume of 0.200 M NaOH required:
Volume of NaOH = Number of moles of NaOH / Concentration of NaOH
Volume of NaOH = 0.00276 mol / 0.200 mol/L = 0.0138 L = 13.8 ml
Therefore, you would need approximately 13.8 ml of 0.200 M NaOH to titrate a 2 ml aliquot of the vinegar.
To find the volume of 0.200 M NaOH needed to titrate a 2 mL sample of vinegar, we first need to determine the number of moles of acetic acid present in the vinegar sample.
Step 1: Calculate the density of acetic acid in the vinegar sample.
Since the vinegar sample is 5.88% acetic acid by volume, we can assume that the remaining 94.12% is water. Therefore, the density of the vinegar sample is 94.12% times the density of water (1.00 g/mL) plus 5.88% times the density of acetic acid (1.409 g/mL).
Density of vinegar sample = (0.9412)(1.00 g/mL) + (0.0588)(1.409 g/mL) = 1.036 g/mL
Step 2: Calculate the mass of acetic acid in the 2 mL vinegar sample.
Mass of acetic acid = volume of vinegar sample (in mL) x density of vinegar sample (in g/mL)
Mass of acetic acid = 2 mL x 1.036 g/mL = 2.072 g
Step 3: Convert the mass of acetic acid to moles.
Molar mass of acetic acid (CH3COOH) = 12.01 g/mol (atomic mass of carbon) + 1.01 g/mol (atomic mass of hydrogen) + (3 x 16.00 g/mol) (sum of atomic mass of oxygen)
Molar mass of acetic acid = 60.05 g/mol
Moles of acetic acid = mass of acetic acid (in grams) / molar mass of acetic acid (in g/mol)
Moles of acetic acid = 2.072 g / 60.05 g/mol = 0.0345 mol
Step 4: Use the balanced chemical equation to determine the mole ratio between acetic acid and NaOH.
The balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:
CH3COOH + NaOH → CH3COONa + H2O
From the equation, we can see that the mole ratio between acetic acid and NaOH is 1:1.
Step 5: Calculate the volume of 0.200 M NaOH needed to react with the moles of acetic acid.
Volume of NaOH (in L) = moles of acetic acid x mole ratio x molar concentration of NaOH (in mol/L)
Volume of NaOH = 0.0345 mol x 1 x 0.200 mol/L = 0.0069 L
Finally, convert the volume of NaOH from liters to milliliters.
Volume of NaOH = 0.0069 L x 1000 mL/L = 6.9 mL
Therefore, a volume of 6.9 mL of 0.200 M NaOH would be required to titrate a 2 mL aliquot of this vinegar.