How do you calculate the pH of acetic acid?

Acetic acid, #"CH"_3"COOH"#, is a weak acid, meaning that it partially ionizes in aqueous solution to form hydronium cations, #"H"_3"O"^(+)#, and acetate anions, #"CH"_3"COO"^(-)#.

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

The position of the ionization equilibrium is given by the acid dissociation constant, #K_a#, which for acetic acid is equal to

#K_a = 1.8 * 10^(-5)#

http://www.bpc.edu/mathscience/chemistry/table_of_monoprotic_acids.html

Now, let's assume that you want to find the pH of a solution of acetic acid that has a concentration of #c#. According to the balanced chemical equation that describes the ionization of the acid, every mole of acetic acid that ionizes will produce

• one mole of hydronium cations
• one mole of acetate anions

If you take #x# to be the concentration of acetic acid that ionizes, you can find the equilibrium concentration of the hydronium cations by using an ICE table

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

#color(purple)("I")color(white)(aaaaaaacolor(black)(c)aaaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaaacolor(black)(0)#
#color(purple)("C")color(white)(aaaacolor(black)((-x))aaaaaaaaaaaaaaaacolor(black)((+x))aaaaaaaacolor(black)((+x))#
#color(purple)("E")color(white)(aaaaacolor(black)(c-x)aaaaaaaaaaaaaaaaaacolor(black)(x)aaaaaaaaaaacolor(black)(x)#

The acid dissociation constant will be equal to

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

This will be equivalent to

#K_(sp) = (x * x)/(c-x) = x^2/(c-x)#

Now, as long as the initial concentration of the acetic acid, #c#, is significantly higher than the #K_(sp)# of the acid, you can use the approximation

#c - x ~~ c -> # valid when #color(red)(ul(color(black)(c " >> " K_(sp)))#

In this case, the equation becomes

#K_(sp) = x^2/c#

which gives you

#x = sqrt(c * K_(sp))#

Since #x# represents the equilibrium concentration of hydronium cations, you will have

#["H"_3"O"^(+)] = sqrt(c * K_(sp))#

Now, the pH of the solution is given by

#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#

Combine these two equations to get

#color(green)(|bar(ul(color(white)(a/a)color(black)("pH" = - log( sqrt(c * K_(sp)))color(white)(a/a)|)))#

For example, the pH of a #"0.050 M"# acetic acid solution will be

#"pH" = - log( 0.050 * 1.8 * 10^(-5))#

#"pH" = 3.02#