Approximately what is the pH of a 0.1 M acetic acid solution?

About #2.87# in aqueous solution.

Well, I assume that you mean in water. I have to assume that, because it matters.

The #"pH"# can be found by finding the #["H"^(+)]# dissociated from acetic acid into water. Therefore, we must write the dissociation reaction to construct the ICE table.

#"HA"(aq) " "+" " "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "A"^(-)(aq)#

#"I"" ""0.1 M"" "" "" "" "-" "" "" ""0 M"" "" "" "" ""0 M"#
#"C"" "-x" "" "" "" "" "-" "" "+x" "" "" "" "+x#
#"E"" "(0.1 - x)"M"" "" "-" "" "" "x" "" "" "" "" "x#

The equilibrium expression is then:

#K_a = x^2/(0.1 - x) = 1.8 xx 10^(-5)#

We assume that for a #K_a# on the order of #10^(-5)# or less, the small #x# approximation works. We can verify this by checking that the percent dissociation is less than #5%# later.

#1.8 xx 10^(-5) ~~ x^2/0.1#

#=> x ~~ sqrt(0.1K_a)#

#= ["H"^(+)] =# #"0.001342 M"#

Therefore, the #"pH"# is:

#color(blue)("pH") ~~ -log["H"^(+)] = color(blue)(2.87)#

And to verify that the percent dissociation is sufficiently small, we check that

#bb(%"dissoc.") = x/(["HA"]) < 0.05#:

#"0.001342 M"/"0.1 M" = bb(0.01342) < 0.05# #color(blue)(sqrt"")#