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

##### 1 Answer

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

The

#"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

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

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

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

Therefore, the

#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"")#