If 20.0 mL of glacial acetic acid (pure HC2H3O2HC2H3O2) is diluted to 1.70 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL.

The answer to your question is pH = 0.686

Explanation:

Data

Acetic acid = 20 ml

Final volume = 1.7 L

pH = ?

density = 1.05 g/ml

Process

1.- Calculate the mass of acetic acid

density = mass / volume

mass = density x volume

mass = 1.05 x 20

mass = 21 g

2.- Calculate the moles of acetic acid (CH₃COOH)

Molar mass = (12 x 2) + (16 x 2) + (4 x 1)

                   = 24 + 32 + 4

                   = 60 g

                 60 g of acetic acid 1 mol

                 21 g of acetic acid x

                 x = (21 x 1) / 60

                 x = 0.35 moles

3.- Calculate the concentration

Molarity = 0.35 moles / 1.70 L

              = 0.206

4.- Calculate the pH

pH = -log[0.206]

pH = 0.686

step-by-step explanation:

whole numbers, zero or greater

answer: i would say b.

explanation: