If 20ml of glacial acetic acid (pure HC2H3O2) is diluted to 1.5L with water, what is the pH of the resulting solution. The density of glacial acetic acid is 1.08g/ml
Call acetic acid HAc.
Calculate M concd HAc.
1.08 g/mL x 1000 mL x (1 mol/molar mass HAc) = ??M HAc.
For the diluted solution,
M HAc x (20 mL/1,500 mL) = ?? M
HAc ==> H^+ + Ac^-
Ka = (H^+)Ac^-)/(HAc)
Set up an ICE chart, substitute into the Ka expression, and solve for H^+, then convert to pH.
СЬфЗ
To determine the pH of the resulting solution, we need to calculate the concentration of the acetic acid in the solution.
First, let's calculate the mass of the glacial acetic acid used.
Mass = Volume x Density
Mass = 20ml x 1.08g/ml
Mass = 21.6g
Next, we convert the mass to moles. The molar mass of acetic acid (HC2H3O2) is 60.05 g/mol.
Moles = Mass / Molar Mass
Moles = 21.6g / 60.05 g/mol
Moles ≈ 0.3599 mol
Since the solution was diluted to 1.5L, we can calculate the molarity of acetic acid.
Molarity = Moles / Volume
Molarity = 0.3599 mol / 1.5 L
Molarity ≈ 0.24 mol/L
Now, we have the molarity of the acetic acid. To determine the pH, we need to consider the dissociation of acetic acid in water. Acetic acid is a weak acid, so it only partially dissociates in water.
The dissociation equation for acetic acid is:
HC2H3O2 ⇌ H+ + C2H3O2-
Since acetic acid is a weak acid, it undergoes only partial dissociation. We can use the concept of the acid dissociation constant (Ka) to determine the concentration of the hydrogen ions (H+) in the solution.
The Ka of acetic acid is approximately 1.8 x 10^-5.
The equation for Ka is:
Ka = [H+][C2H3O2-] / [HC2H3O2]
In an acidic solution, the concentration of [H+] is the same as the concentration of the acetic acid that dissociates, since 1 mole of acetic acid dissociates into 1 mole of H+ ions.
Since acetic acid partially dissociates, we can assume that the initial concentration of acetic acid is equal to its final concentration after dissociation.
Thus, [HC2H3O2] = 0.24 mol/L.
Using the value of Ka, we can calculate the concentration of H+ ions.
[H+] = sqrt(Ka x [HC2H3O2])
[H+] = sqrt((1.8 x 10^-5) x (0.24 mol/L))
[H+] ≈ 2.433 x 10^-3 mol/L
Finally, we can calculate the pH of the resulting solution using the concentration of H+ ions.
pH = -log[H+]
pH = -log(2.433 x 10^-3 mol/L)
pH ≈ 2.613
Therefore, the pH of the resulting solution is approximately 2.613.