If you assume that there is 5% acetic acid in vinegar, how much vinegar should you weigh out so that the endpoint requires 25.0 mL of 0.186 M NaOH?
Is that 5% w/w or some other unit? I will assume w/w. That means 5g CH3COOH in 100 g soln. Convert 5g to mols. That's 5/molar mass = about 0.083 mols. How many mols do you need? That equals to mols NaOH = M x L = 0.025 x 0.186 = about 0.0046. So how much of the 0.083 mol vinegar to weigh. It is
0.083mols x (?g/100g) = 0.0046
Solve for ?g.
5.58 g
To find out how much vinegar to weigh out, we need to calculate the moles of acetic acid present in the vinegar solution.
Step 1: Calculate the moles of NaOH used.
Using the equation:
Moles of NaOH = Volume (L) × Concentration (M)
Moles of NaOH = 0.0250 L × 0.186 M
Moles of NaOH = 0.00465 moles
Step 2: Calculate the moles of acetic acid.
Since acetic acid (CH3COOH) and NaOH react in a 1:1 ratio, the moles of acetic acid are also 0.00465 moles.
Step 3: Convert moles of acetic acid to grams.
The molar mass of acetic acid is 60.05 g/mol.
Mass of acetic acid = Moles × Molar mass
Mass of acetic acid = 0.00465 moles × 60.05 g/mol
Mass of acetic acid = 0.279 g
Step 4: Calculate the mass of vinegar needed.
Since vinegar is a 5% solution of acetic acid, we can assume that every 100 g of vinegar contains 5 g of acetic acid.
By setting up a proportion, we can find the mass of vinegar needed.
(0.279 g acetic acid) / (5 g vinegar) = (x g acetic acid) / (100 g vinegar)
Cross-multiplying, we find:
0.279 g acetic acid = (5 g vinegar * x g acetic acid) / 100 g vinegar
0.279 g acetic acid = 0.05x
x = (0.279 g acetic acid) / 0.05
x ≈ 5.58 g
Therefore, you should weigh out approximately 5.58 grams of vinegar to obtain the desired endpoint with 25.0 mL of 0.186 M NaOH.
To determine how much vinegar you should weigh out, you need to calculate the amount of acetic acid present in the vinegar and then convert it to grams.
1. Start by determining the moles of acetic acid required to neutralize the NaOH. Use the balanced chemical equation:
CH3COOH + NaOH → CH3COONa + H2O
From the equation, it is clear that the stoichiometric ratio between acetic acid and NaOH is 1:1.
The amount of NaOH used is given as 25.0 mL. Convert this to liters by dividing by 1000:
25.0 mL ÷ 1000 = 0.025 L
Now, calculate the moles of acetic acid:
Moles of NaOH = Volume (L) × Concentration (mol/L)
= 0.025 L × 0.186 mol/L
= 0.00465 mol
2. Since the stoichiometric ratio between acetic acid and NaOH is 1:1, the moles of acetic acid are also 0.00465 mol.
3. Now, determine the mass of acetic acid using the molar mass of acetic acid, which is 60.05 g/mol. Multiply the moles of acetic acid by its molar mass:
Mass of acetic acid = Moles of acetic acid × Molar mass
= 0.00465 mol × 60.05 g/mol
= 0.279 g
4. Since the concentration of acetic acid in vinegar is assumed to be 5%, we can calculate the mass of vinegar needed:
Mass of vinegar = Mass of acetic acid ÷ Percent concentration
= 0.279 g ÷ 0.05
= 5.58 g
Therefore, you should weigh out approximately 5.58 grams of vinegar to obtain the desired amount of acetic acid for the endpoint requiring 25.0 mL of 0.186 M NaOH.