100.0 mL Vinegar (d=1.06g/mL) contains 4% acetic acid.
a. How many moles of acid are in 100 mL of vinegar?
b. How many mL of 0.11 M NaOH are required to neutralize the vinegar?
That is super helpful, thanks. I missed a two days of class when we learned acids and bases, so I was a bit stuck.
To calculate the number of moles of acid in 100 mL of vinegar, we first need to find the mass of the acetic acid present in the solution.
a. The concentration of acetic acid in the vinegar solution is given as 4%, which means that 4 grams of acetic acid are present in every 100 mL of vinegar.
To find the mass of acetic acid, we multiply the volume (100 mL) by the density (1.06 g/mL) of vinegar:
Mass = Volume x Density
Mass = 100 mL x 1.06 g/mL
Mass = 106 g
Since the concentration of acetic acid is 4%, we can calculate the mass of acetic acid present in the solution:
Mass of acetic acid = (4/100) x Mass
Mass of acetic acid = (4/100) x 106 g
Mass of acetic acid = 4.24 g
The molar mass of acetic acid (CH3COOH) is 60.05 g/mol. To find the number of moles of acetic acid, we divide the mass of acetic acid by its molar mass:
Number of moles = Mass of acetic acid / Molar mass
Number of moles = 4.24 g / 60.05 g/mol
Number of moles = 0.0706 mol
Therefore, there are approximately 0.0706 moles of acetic acid in 100 mL of vinegar.
b. To find the volume of 0.11 M NaOH required to neutralize the vinegar, we can use the balanced equation between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH -> CH3COONa + H2O
From the equation, we can see that one mole of acetic acid reacts with one mole of NaOH.
Using the mole ratio, we can now calculate the volume of NaOH solution needed to neutralize the acetic acid.
1 mole of acetic acid reacts with 1 mole of NaOH
Therefore, 0.0706 moles of acetic acid will react with the same number of moles of NaOH:
Volume of NaOH solution = Number of moles of acetic acid / Molarity of NaOH solution
Volume of NaOH solution = 0.0706 mol / 0.11 mol/L
Volume of NaOH solution = 0.641 L
Since the molarity of the NaOH solution is given in liters, we need to convert it to milliliters:
Volume of NaOH solution = 0.641 L x 1000 mL/L
Volume of NaOH solution = 641 mL
Therefore, approximately 641 mL of 0.11 M NaOH solution is required to neutralize 100 mL of vinegar.
To answer these questions, we need to use the given information about the concentration of acetic acid in vinegar and the density of vinegar.
a. To calculate the number of moles of acid in 100 mL of vinegar, we first need to determine the mass of acetic acid present in 100 mL.
Given that the density of vinegar (d) is 1.06 g/mL, we can calculate the mass of 100 mL of vinegar as follows:
Mass = Volume × Density = 100 mL × 1.06 g/mL = 106 g
Since the vinegar contains 4% acetic acid, we can calculate the mass of acetic acid in 100 mL of vinegar:
Mass of acetic acid = 4% × Mass of vinegar = 4/100 × 106 g = 4.24 g
Next, we need to calculate the molar mass of acetic acid. The molar mass of acetic acid (CH3COOH) is approximately 60.052 g/mol.
Using the molar mass, we can calculate the number of moles of acetic acid in 4.24 g:
Number of moles = Mass / Molar mass = 4.24 g / 60.052 g/mol ≈ 0.0706 mol
Therefore, there are approximately 0.0706 moles of acetic acid in 100 mL of vinegar.
b. To calculate the volume of 0.11 M NaOH required to neutralize the vinegar, we need to use the following balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH):
CH3COOH + NaOH → CH3COONa + H2O
From the balanced equation, we can see that the stoichiometric ratio between acetic acid and sodium hydroxide is 1:1.
Since we found that there are 0.0706 moles of acetic acid in 100 mL of vinegar, we know that we need an equal number of moles of NaOH for complete neutralization.
Therefore, the number of moles of NaOH required is also 0.0706 mol.
Now we can use the molarity and mole ratio to calculate the required volume of 0.11 M NaOH.
Molarity (M) is defined as moles of solute per liter of solution.
Moles = Molarity × Volume (in liters)
Volume = Moles / Molarity
Using this formula, we can calculate the volume of 0.11 M NaOH:
Volume = 0.0706 mol / 0.11 mol/L = 0.641 L
To convert this volume from liters to milliliters, we multiply by 1000:
Volume = 0.641 L × 1000 mL/L ≈ 641 mL
Therefore, approximately 641 mL of 0.11 M NaOH are required to neutralize 100 mL of vinegar.
Is that 4% w/w. If so, that means 4g acetic acid/100 g soln.
4 g acetic acid = 4/molar mass = approximately 0.7 mols (but you need to do that exactly).
Convert 100 g soln to L, then M vinegar = mols/L.
For part b, mL NaOH x M NaOH = mL vinegar x ZM vinegar.