4. 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?
How many moles NaOH did you use? That is M x L = ??
How many moles vinegar do you have in the 10 mL? Since the reaction is 1:1 (see equation below) the same number of moles of vinegar are in the 10 mL sample.
Concn = M = moles/L. So that many moles in 0.010 L = ??
CH3COOH + NaOH ==> CH3COONa + H2O
To determine the concentration of acetic acid in the vinegar, we can use the concept of stoichiometry and the information given in the question.
First, let's write a balanced chemical equation for the neutralization reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH → CH3COONa + H2O
Based on the balanced equation, we can see that one mole of acetic acid reacts with one mole of sodium hydroxide to produce one mole of sodium acetate and one mole of water.
Now, let's calculate the number of moles of NaOH used in the neutralization reaction:
moles of NaOH = volume of NaOH (in liters) * concentration of NaOH
Given:
- Volume of NaOH = 26.5 mL = 0.0265 L
- Concentration of NaOH = 0.250 M
moles of NaOH = 0.0265 L * 0.250 M = 0.006625 moles
Since the balanced equation shows that the ratio of moles of acetic acid to moles of NaOH is 1:1, we know that the number of moles of acetic acid in 10.0 mL of vinegar is also 0.006625 moles.
Now, let's calculate the concentration of acetic acid in the vinegar:
concentration of acetic acid = moles of acetic acid/volume of vinegar (in liters)
Given:
- Volume of vinegar = 10.0 mL = 0.01 L
concentration of acetic acid = 0.006625 moles / 0.01 L = 0.6625 M
Therefore, the concentration of acetic acid in the vinegar is 0.6625 M.