Question

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

Answer 1

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"")`