A total of 31.3 ml of 0.15 N NaOH was required to reach a phenolphthalein endpoint in titrating 6.00 g of vinegar. Calculate the number of equivalents of acetic acid in vinegar.
mL x N = # millieqivalents
Divide by 1000 to convert to eq.
You don't need the 6.00 g vinegar unless somewhere down the line you need grams acetic acid in the vinegar.
To calculate the number of equivalents of acetic acid in vinegar, we need to first find the moles of acetic acid in vinegar using the equation:
moles of acetic acid = volume (in liters) * concentration (in moles per liter)
Given:
Volume of NaOH = 31.3 mL = 0.0313 L
Concentration of NaOH = 0.15 N
To find the concentration in moles per liter (M), we need to convert the Normality (N) of NaOH to Molarity (M). The conversion is as follows:
M = N * molar mass of solute
The molar mass of NaOH is 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 39.99 g/mol.
Concentration of NaOH in Molarity (M) = 0.15 N * 39.99 g/mol = 5.9985 M
Now we can calculate the moles of NaOH:
moles of NaOH = volume (in liters) * concentration (in moles per liter)
= 0.0313 L * 5.9985 M
= 0.1868 moles of NaOH
Since the reaction between NaOH and acetic acid is 1:1, the moles of acetic acid in vinegar will also be 0.1868 moles.
Now, calculate the molar mass of acetic acid (CH3COOH):
molar mass of acetic acid = (12.01 g/mol * 2) + (1.01 g/mol * 3) + 16.00 g/mol + 1.01 g/mol
= 60.05 g/mol
Finally, we can calculate the number of equivalents of acetic acid using the equation:
equivalents of acetic acid = moles of acetic acid / valence factor
The valence factor for acetic acid is 1, since it donates 1 hydrogen ion in the reaction.
equivalents of acetic acid = 0.1868 moles / 1
= 0.1868 equivalents
Therefore, there are 0.1868 equivalents of acetic acid in 6.00 g of vinegar.