A 50ml sample of vinegar is titrated with 0.774M NaOH(aq). If the titration requires 41.6ml of NaOH(aq), what is the concentration of acetic acid in the vinegar?

To find the concentration of acetic acid in the vinegar, we need to use the concept of stoichiometry and the equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):

CH3COOH + NaOH -> CH3COONa + H2O

First, let's find the number of moles of NaOH used in the titration:

Number of moles of NaOH = (volume of NaOH used in titration) x (molarity of NaOH)
= 41.6 ml x 0.774 M
= 32.1184 mmol

Since the equation shows that the stoichiometric ratio between NaOH and CH3COOH is 1:1, the number of moles of acetic acid (CH3COOH) present in the vinegar sample is also 32.1184 mmol.

Next, we need to convert the volume of vinegar from milliliters to liters:

Volume of vinegar = 50 ml = 50 / 1000 L = 0.05 L

Now, we can calculate the concentration of acetic acid in the vinegar:

Concentration of acetic acid = (number of moles of acetic acid) / (volume of vinegar)
= 32.1184 mmol / 0.05 L
= 642.368 M

Therefore, the concentration of acetic acid in the vinegar is 642.368 M.

mL acid x M acid = mL base x M base.