a 10.0mL vinegar sample was completely neutralized by 22.5mL 0.2M NaOH solution. calculate molarity and % of acetic acid in vinegar.
please help
I assume we call the density of the vinegar 1.00 g/mL so the 10 mL has a mass of 10.0 grams.
moles NaOH used = M x L = ??
moles of acetic acid in the 10.0 mL = ?? (same as moles NaOH).
M acetic acid = moles/L = ??/0.1 = xx M.
Is that percent w/w or w/v. I assume w/v.
You have M and that is moles/L.
moles x molar mass = grams so that is g/L
You want g/100 mL = ??
i'm still lost can you show me step by step please
To solve this problem, we need to use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH). The balanced equation is as follows:
CH3COOH + NaOH → CH3COONa + H2O
From the balanced equation, we can see that the mole ratio between CH3COOH and NaOH is 1:1. This means that for every 1 mole of CH3COOH, 1 mole of NaOH is required for complete neutralization.
Step 1: Calculate the moles of NaOH used.
Given: Volume of NaOH solution = 22.5 mL = 0.0225 L
Concentration of NaOH solution = 0.2 M
Moles of NaOH = Volume (in L) x Concentration
= 0.0225 L x 0.2 M
= 0.0045 moles
Step 2: Determine the moles of CH3COOH.
Since the mole ratio between CH3COOH and NaOH is 1:1, the moles of CH3COOH will also be 0.0045 moles.
Step 3: Calculate the molarity of CH3COOH.
Molarity is defined as moles of solute per liter of solution.
Volume of vinegar sample = 10.0 mL = 0.010 L
Molarity of CH3COOH = Moles of CH3COOH / Volume of CH3COOH (in L)
= 0.0045 moles / 0.010 L
= 0.45 M
Therefore, the molarity of acetic acid (CH3COOH) in the vinegar sample is 0.45 M.
Step 4: Calculate the percent of acetic acid in vinegar.
To calculate the percent composition, we need to compare the moles of CH3COOH with the total mass of the vinegar sample.
The molar mass of CH3COOH (acetic acid) = 60.05 g/mol
Mass of CH3COOH = Moles of CH3COOH x Molar mass of CH3COOH
= 0.0045 moles x 60.05 g/mol
= 0.2702 g
Mass of vinegar sample = Volume of vinegar sample x Density of vinegar
= 10.0 mL x (vinegar density in g/mL) (You may need to look up the density of vinegar from a reliable source)
% of acetic acid in vinegar = (Mass of CH3COOH / Mass of vinegar sample) x 100
= (0.2702 g / Mass of vinegar sample) x 100
To get the % of acetic acid in vinegar, you need to know the mass of the vinegar sample and its density. Once you have that information, you can substitute it into the equation above to get the answer.
(Note: The answer will vary depending on the mass of the vinegar sample and its density.
I worked the problem for you. All you need to do is substitute a few numbers. Post your work and tell me what you don't understand at the next step and I'll try to help you through BUT I am not going to do the problem for you.
is the answer M= .08M
and is the % of mass= 44%