Vinegar is a dilute solution of acetic acid. What is the concentration of acetic acid in the vinegar if 26.5 mL of NaOH at 0.250 M are required to neutralize 10.0 mL of vinegar?

Write the equation. It tells you that 1 mole NaOH = 1 mole acetic acid.

CH3COOH + NaOH ==> CH3COONa + H2O

moles NaOH = M x L = ??
Since the ratio of acetic acid to NaOH is 1:1, then moles acetic acid = moles NaOH.

Finally, M acetic acid = moles acetic acid/L acetic acid. You know moles and L acetic acid, solve for M.

To find the concentration of acetic acid in the vinegar, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between acetic acid and sodium hydroxide.

The balanced chemical equation for the neutralization reaction is:

CH3COOH (acetic acid) + NaOH (sodium hydroxide) → CH3COONa (sodium acetate) + H2O (water)

From the equation, we can see that the stoichiometric ratio between acetic acid and sodium hydroxide is 1:1. This means that for every 1 mole of acetic acid, 1 mole of sodium hydroxide is required to neutralize it.

Now, let's calculate the number of moles of NaOH used in the reaction:

moles of NaOH = concentration of NaOH × volume of NaOH (in liters)
= 0.250 M × (26.5 mL / 1000 mL)
= 0.006625 moles

Since the stoichiometric ratio between acetic acid and NaOH is 1:1, the number of moles of acetic acid present in 10.0 mL of vinegar is also 0.006625 moles.

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

concentration of acetic acid = moles of acetic acid / volume of vinegar (in liters)
= 0.006625 moles / (10.0 mL / 1000 mL)
= 0.6625 M

Therefore, the concentration of acetic acid in the vinegar is 0.6625 M.

To find the concentration of acetic acid in the vinegar, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between acetic acid and sodium hydroxide.

The balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:

CH3COOH + NaOH -> CH3COONa + H2O

From the balanced equation, we can see that the stoichiometric ratio between acetic acid and sodium hydroxide is 1:1. This means that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide.

Given that 26.5 mL of 0.250 M NaOH is required to neutralize 10.0 mL of vinegar, we can calculate the number of moles of NaOH used.

moles of NaOH = volume of NaOH (in L) * molarity of NaOH
= 26.5 mL * (1 L / 1000 mL) * 0.250 mol/L
= 0.006625 mol

Since the stoichiometric ratio between acetic acid and sodium hydroxide is 1:1, the number of moles of acetic acid in the vinegar is also 0.006625 mol.

Now, let's calculate the concentration of acetic acid in the vinegar.

concentration of acetic acid = moles of acetic acid / volume of vinegar (in L)
= 0.006625 mol / (10.0 mL * (1 L / 1000 mL))
= 0.6625 mol/L

Therefore, the concentration of acetic acid in the vinegar is 0.6625 mol/L.