If #15.0mL# of glacial acetic acid (pure #HC_2H_3O_2#) is diluted to #1.50L# with water, what is the pH of the resulting solution? The density of glacial acetic acid is #"1.05 g/mL"#

To be able to solve this problem, you need to know the value of acetic acid's acid dissociation constant, #K_a#, which essentially tells you how how many acid molecules will ionize in solution.

The acid dissociation constant for acetic acid is listed as

#K_a = 1.76 * 10^(-5)#

So, the idea is that you dilute a sample of pure acetic acid by a dilution factor of 100. The density of the pure acetic acid will help you determine how many grams of acid you're diluting.

#15.0color(red)(cancel(color(black)("mL"))) * "1.05 g"/(1color(red)(cancel(color(black)("mL")))) = "15.75 g HAc"#

Use the acid's molar mass to see how many moles you'd get in that much mass

#15.75color(red)(cancel(color(black)("g"))) * "1 mole HAc"/(60.05color(red)(cancel(color(black)("g")))) = "0.2623 moles HAc"#

This means that the acetic acid solution has a molarity of

#C = n/V = "0.2623 moles"/("1.50 L") = "0.175 M"#

Next, use an ICE Table to figure out exactly how many hydronium ions you'd get from the acid's dissociation.

#" "CH_3COOH_((aq)) + H_2O_((l)) rightleftharpoons H_3O_((l))^(+) + CH_3COO_((aq))^(-)#
I............0.175............................................0....................0
C............(-x)..............................................(+x)...............(+x)
E..........0.175-x..........................................x...................x

By definition, #K_a# will be

#K_a = ([H_3O^(+)] * [CH_3COO^(-)])/([CH_3COOH])#

#K_a = (x * x)/(0.175 - x) = 1.76 * 10^(-5)#

Because #K_a# is so small, you can approximate 0.175 - x with 0.175 to get

#K_a = x^2/0.175 = 1.76 * 10^(-5)#

#x = sqrt(3.08 * 10^(-6)) = 1.75 * 10^(-3)#

The concentration of the hydronium ions will thus be

#x = [H_3O^(+)] = 1.75 * 10^(-3)"M"#

The pH of the solution will be equal to

#pH_"sol" = -log([H_3O^(+)])#

#pH_"sol" = -log(1.75 * 10^(-3)) = color(green)(2.757)#