Question

If 15.0 mL of glacial acetic acid (pure HC2H3O2) 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


pH=?

Answer 1

Mass of CH3COOHvolume x density = 15.0 x 1.05 = 15.75 g Initial moles of CH3COOH mass/molar mass of CH3COOH 15.75/60.05 0.2623 mol Initial concentration of CH3COOH = moles/volume 0.2623/1.70 0.1543 M CH3COOH CH3COO-+H+ > 0.1543 +a +a 0.1543-а Ka = [CH3COO-] [H+] /[CH3COOH] a (0.1543 a) 1.8 x 10 a2 +1.8 x 10-a 2.777 x 10-6 0 The positive root of the quadratic equation is: a 0.001657 [H+]- a 0.001657 M pH log[H+log(0.001657) 2.78