If 20.0 mL of glacial acetic acid (pure HC2H3O2) is diluted to 1.30 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL.
How much will 1 L of glacial acetic acid weigh? 1000 mL x 1.05 g/mL = 1050 grams.
How much of that is acetic acid. If it's glacial acetic acid, it is 100% acetic acid so all 1050 g is HC2H3O2.
How many moles is that?
1050 grams/molar mass = 1050/60 = 17.50 M.
The new concn will be 17.5 M x (20 mL/1300 mL) = ?? M (or you can do a mL x M = mL x M and come up with the same thing).
Next the ionization.
HAc (easier to write than HC2H3O2) + H2O ==> H3O^+ + Ac^-
Ka = (H^+)(Ac^-)/(HAc) = 1.8 x 10^-5
Set up an ICE chart, calculate H^+, then convert to pH. Check my thinking.
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To find the pH of the resulting solution, we first need to calculate the concentration of acetic acid (HC2H3O2) in the solution.
Step 1: Calculate the mass of glacial acetic acid used.
The density of glacial acetic acid is given as 1.05 g/mL. Therefore, the mass of 20.0 mL of glacial acetic acid can be calculated as:
Mass = Volume x Density
Mass = 20.0 mL x 1.05 g/mL
Mass = 21.0 g
Step 2: Calculate the molarity of acetic acid in the solution.
Molarity (M) is defined as moles of solute per liter of solution. Since 20.0 mL is diluted to 1.30 L, the volume of the resulting solution is 1.30 L.
Moles of acetic acid = mass / molar mass
Molar mass of acetic acid (HC2H3O2) = 2(1.01) + 4(12.01) + 2(16.00) = 60.05 g/mol
Moles of acetic acid = 21.0 g / 60.05 g/mol = 0.349 mol
Molarity (M) = Moles / Volume
Molarity (M) = 0.349 mol / 1.30 L
Molarity (M) = 0.268 M
Step 3: Calculate the concentration of H+ ions (protons) in the solution.
Since acetic acid is a weak acid, it partially dissociates in water to form H+ ions and the acetate ion (C2H3O2-). The balanced equation for the dissociation is:
HC2H3O2 ⇌ H+ + C2H3O2-
In this case, the concentration of H+ ions is equal to the concentration of acetic acid, which is 0.268 M.
Step 4: Calculate the pOH of the solution.
pOH = -log10(OH-)
Since we have the concentration of H+ ions, we can calculate OH- concentration using the equation:
[H+] x [OH-] = 1 x 10^-14 (for water at 25°C)
[H+] x [OH-] = Kw
where Kw is the ion product constant for water, which is equal to 1 x 10^-14.
[OH-] = Kw / [H+]
[OH-] = 1 x 10^-14 / 0.268 M
[OH-] = 3.731 x 10^-14 M
pOH = -log10(3.731 x 10^-14 M)
pOH ≈ 13.43
Step 5: Calculate the pH of the solution.
Since pH + pOH = 14 (at 25°C), we can calculate the pH as:
pH = 14 - pOH
pH = 14 - 13.43
pH ≈ 0.57
Therefore, the pH of the resulting solution is approximately 0.57.