Question

What is the molarity of the sodium hydroxide solution used in the titration? 0.4750 M In the space below, type in the balanced molecular equation between acetic acid and sodium hydroxide (Make sure to use subscripts) CH3CO2H (aq) NaOH (aq)CH3CO2Na (aq) H20 (aq) What was the average volume of sodium hydroxide (report the average that included the more reproducible volumes from your ttration)? 17.30 m. What was the % rsd you calculated using these volumes?,040 %- Calculate the molarity of the acetic acid contained in 10.00 mL of vinegar using the molanity...and the average volume of sodium hydroxide and the reaction mentioned above. You can handwrite this part in the space below (not on a separate piece of paper). Clearly show ALL your calculations What is the mass of acetic acid contained in 10.00 mL of vinegar? You can handwrite this part in the space below (not on a separate piece of paper). Clearly show ALL your calculations

Answer 1

To calculate the molarity of acetic acid, the moles of NaOH used must be calculated:

n NaOH = 0.475 mol / L * 17.30 mL * (1 L / 1000 mL) = 8.22x10 ^ -3 moles

One mole of NaOH equals one mole of acetic acid, then you have the same number of moles of acetic acid and the molarity is calculated:

M = 8.22x10 ^ -3 moles / 0.01 L = 0.822 M

The mass of acetic acid is calculated from the moles of acetic acid:

mass AA = 8.22x10 ^ -3 moles * (60.05 g / 1 mol) = 0.4936 g

the mass equivalent to the solution is calculated:

Mass Sol = 10 mL * (1.018 g / 1 mL) = 10.18 g

The percentage of AA in solution is calculated:

% AA = 0.4936 g * 100 / 10.18 g = 4.85%.