what sample size of .9181M vinegar would require 27.50 mL of .5185M of NaOH for neutralization?

How many moles NaOH do you have? That is M x L = ??

How many moles vinegar do you have? That will be the same as the number of moles NaOH since the equation is a 1:1 ratio.
CH3COOH + NaOH ==> H2O + CH3COONa
Then M vin = moles vin/L vin and solve for L vinegar.

Thanks Dr.Bob!

To determine the sample size of the vinegar, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between vinegar (acetic acid) and sodium hydroxide (NaOH). The balanced equation is:

CH3COOH + NaOH → CH3COONa + H2O

From the equation, we can see that 1 mole of acetic acid (CH3COOH) reacts with 1 mole of NaOH. Therefore, the moles of acetic acid in the vinegar sample can be determined using the following formula:

moles of acetic acid = (volume of NaOH solution in liters) × (molarity of NaOH)

Given that the volume of NaOH solution used is 27.50 mL (which is equal to 0.02750 L) and the molarity of NaOH is 0.5185 M, we can calculate the moles of acetic acid reacted.

moles of acetic acid = 0.02750 L × 0.5185 M = 0.01426 moles

Since the stoichiometry of the reaction tells us that 1 mole of acetic acid reacts with 1 mole of NaOH, the moles of acetic acid are also equal to the moles of NaOH reacted.

Next, we can calculate the number of moles of vinegar used, given that its concentration is 0.9181 M (Molarity).

moles of vinegar = moles of acetic acid = 0.01426 moles

Finally, we need to convert the moles of vinegar to its sample size. To do this, we use the formula:

mass of vinegar = moles of vinegar × molar mass of acetic acid

The molar mass of acetic acid (CH3COOH) is calculated as follows:

(molar mass of carbon) + 3 × (molar mass of hydrogen) + (molar mass of oxygen) + 2 × (molar mass of oxygen)
= (12.01 g/mol) + 3 × (1.01 g/mol) + (16.00 g/mol) + 2 × (16.00 g/mol)
= 60.05 g/mol

Now, we can calculate the mass of vinegar:

mass of vinegar = 0.01426 moles × 60.05 g/mol = 0.857 g

Therefore, a sample size of 0.857 grams of vinegar would require 27.50 mL of 0.5185 M NaOH for neutralization.