If 20.0 ml of glacial acetic acid (pure hc2h3o2) is diluted to 1.40 l with water, what is the ph of the resulting solution? the density of glacial acetic acid is 1.05 g/ml.

By using the formula, mass = density x volume, we calculate mass in grams

 20.0 mL CH₃COOH x (1.05 g / mL) = 21.0 g CH₃COOH 

To find the moles, molar mass of CH₃COOH = 60.05g/mol

21.0 g CH₃COOH x (1 mole CH₃COOH / 60.05 g CH₃COOH) = 0.350 moles CH₃COOH 

To find molarity,

[CH₃COOH] = moles CH₃COOH / L of solution = 0.350 / 1.40 = 0.250 M CH₃COOH 

When CH₃COOH is dissolved in water, it produces small and equal amounts of H₃O⁺+ and C₂H₃O₂⁻. 

Molarity ,         CH₃COOH + H₂O <==> H₃O⁺ + C₂H₃O₂⁻ 

Initial                      0.250                          0           0 
Change                      -x                            x            x 
Equilibrium            0.250-x                        x            x 

Kₐ = [H₃O⁺][C₂H₃O₂⁻] / [HC₂H₃O₂] = (x)(x) / (0.250-x) = 1.8 x 10⁻⁵

Since Kₐ is relatively small, we can neglect the -x term after 0.250 to simplify 

x² / 0.250 = 1.8 x 10⁻⁵ 

x² = 4.5 x 10⁻⁶ 

x = 2.1 x 10⁻³ = [H₃O⁺] 

pH = -log [H₃O⁺] = -log (2.1 x 10⁻³) = 2.68

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