Question

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. Express your answer to two decimal places. pH =

Answer 1

Mass of acetic acid = volume * density = 20.0 mL * 1.05 g/mL 21 g Molecula r weight of acetic acid = 60.05 g/mol Number of moles of acetic acid -21 g/(60.05 g/mol) 0.349709 mol Volume of the solution 1.30 L Concentration of acetic acid 0.349709 mol/1.30 L - 0.269007 M Let x be the change in the conc. of each species at equilibrium CH3COOH + H2O-_ CH3COO initial 0.26901 0 0 change + X Equilibrium 0.26901 -x x..x =1.8x10-5 0.2690 1-X r2 +1.8x10r-4.84x10-60 by solving, we get x-一0.00221 and 0.002192 r cannot be negative, so, x-0.002192 pH--log[Hllog (0.002192) 2.66