Commercial vinegar usually has a concentration of 5%, per volume.
a) Calculate the [H3O+] , if the density of acetic acid is 1.051 g/mL.
b) Calculate the pH.
5% w/v means 5 g acetic acid/100 mL soln.
You can convert to M by
5g/molar mass = mols.
mols/L soln = M
You need a K to calculate H^+.
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To calculate the concentration of H3O+ (hydronium ion) and the pH of commercial vinegar, we need to use the given information about the acetic acid density and concentration.
a) Calculate the [H3O+]:
To determine the concentration of [H3O+], we need to convert the density to molarity.
First, we need to convert the density from grams per milliliter (g/mL) to grams per liter (g/L) by multiplying it by 1000 since there are 1000 mL in 1 L:
Density = 1.051 g/mL
Density = 1.051 * 1000 g/L
Density = 1051 g/L
Next, we need to convert the grams of acetic acid to moles. The molar mass of acetic acid (CH3COOH) is approximately 60.05 g/mol.
Moles of acetic acid = Mass of acetic acid / Molar mass of acetic acid
Moles of acetic acid = 1051 g / 60.05 g/mol
Moles of acetic acid ≈ 17.5 mol
Since vinegar has a concentration of 5%, per volume, this means there are 5 grams of acetic acid for every 100 mL of vinegar. We can use this information to calculate the volume of vinegar needed to obtain the 17.5 moles of acetic acid:
(5 grams / 100 mL) * V = 1051 g
V = (1051 g / 5 grams) * 100 mL
V ≈ 21.02 L
Now, we have the volume of vinegar needed to obtain 17.5 moles of acetic acid. We can convert this to liters:
V = 21.02 L
Finally, we can calculate the concentration of [H3O+] by dividing the number of moles of acetic acid by the volume of vinegar:
[H3O+] = Moles of acetic acid / Volume of vinegar
[H3O+] = 17.5 mol / 21.02 L
[H3O+] ≈ 0.833 mol/L
Therefore, the concentration of [H3O+] in commercial vinegar is approximately 0.833 mol/L.
b) Calculate the pH:
The pH is a measure of the acidity of a solution and is defined as the negative logarithm of the hydronium ion concentration ([H3O+]).
pH = -log[H3O+]
pH = -log(0.833)
pH ≈ 0.08
Therefore, the pH of commercial vinegar is approximately 0.08.