#"20.0 mL"# of glacial acetic acid (pure #"CH"_3"COOH"#) is diluted to #"1.30 L"# with water. The density of glacial acetic acid is #"1.05 g/mL"#. What is the pH of the resulting solution?

Start by using the density of glacial acetic acid to calculate the mass of acetic acid present in the solution.

#20.0 color(red)(cancel(color(black)("mL"))) * "1.05 g"/(1color(red)(cancel(color(black)("mL")))) = "21.0 g"#

Next, use the molar mass of acetic acid to calculate the number of moles present in the solution.

#21.0 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COOH")/(60.05color(red)(cancel(color(black)("g")))) = "0.3497 moles CH"_3"COOH"#

The total volume of the solution is equal to #"1.30 L"#, so use it to find the molarity of the acid.

#["CH"_3"COOH"] = "0.3497 moles"/"1.30 L" = "0.269 M"#

Now, acetic acid is a weak acid, which implies that it will only partially ionize in aqueous solution.

#"CH"_ 3"COOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)#

By definition, the acid dissociation constant, #K_a#, is equal to

#K_a = (["CH"_3"COO"^(-)] * ["H"_ 3"O"^(+)])/(["CH"_3"COOH"])#

Notice that every mole of acetic acid that dissociates produces #1# mole of acetate anions and #1# mole of hydronium cations. This means that if you take #x# #"M"# to be the concentration of acetic acid that dissociates, you can say that, at equilibrium, the solution will contain

#["CH"_ 3"COO"^(-)] = ["H"_ 3"O"^(+)] = x quad "M"#

Similarly, the equilibrium concentration of the acetic acid will be

#["CH"_3"COOH"] = (0.269 - x) quad "M"#

This means that when #x# #"M"# dissociates, the initial concentration of the acid decreases by #x# #"M"#.

Pug this into the expression of the acid dissociation constant to get

#K_a = (x * x)/(0.269 - x)#

#K_a = x^2/(0.269 - x)#

Now, you didn't provide a value for the acid dissociation constant, but I know from experience that this value is small enough to allow you to use the following approximation.

#0.269 - x ~~ 0.269#

This means that you have

#K_a = x^2/0.269#

which gets you

#x = sqrt(0.269 * K_a)#

Since we've said that #x# #"M"# represents the equilibrium concentration of the hydronium cations (and of the acetate anions), you will have

#["H"_ 3"O"^(+)] = sqrt(0.269 * K_a) quad "M"#

Finally, the #"pH"# of the solution is given by

#"pH" = - log(["H"_3"O"^(+)])#

In your case, this is equal to

#"pH" = - log(sqrt(0.269 * K_a))#

Now all you have to do is to use the value of the acid dissociation constant given to you--or do a quick search for the acid dissociation constant of acetic acid--and plug it into the equation.

The value should be rounded to three decimal places because you have three sig figs for your values.