Question

Question #6b487

Answer 1

`"0.55 M"`

Explanation:

Start by calculating the concentration of hydronium cations needed to produce a `"pH"` equal to that of gastric juice.

You know that

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

so

`["H"_3"O"^(+)] = 10^(-"pH")`

In your case, you have

`["H"_3"O"^(+)] = 10^(-2.50) = 3.16 * 10^(-3)` `"M"`

Now, acetic acid is a weak acid, which implies that it does not ionize completely in aqueous solution. Instead, an equilibrium is established between the unionized acetic acid molecules and the ions

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

Notice that each mole of acetic acid that ionizes produces `1` mole of acetate anions and `1` mole of hydronium cations.

This means that, at equilibrium, you have

`["CH"_3"COO"^(-)] = ["H"_3"O"^(+)]`

If you take `x` `"M"` to be the initial concentration of the acid, you can say that, at equilibrium, you will have

`["CH"_3"COOH"] = x - ["H"_3"O"^(+)]`

The concentration of the acid decreases by the concentration that ionizes.

By definition, the acid dissociation constant, `K_a`, for this equilibrium looks like this

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

In your case, you have

`1.8 * 10^(-5) = (3.16 * 10^(-3) * 3.16 * 10^(-3))/(x - 3.16 * 10^(-3))`

Now, because you have such a small value for `K_a`, you can assume that

`x - 3.16 * 10^(-3) ~~ x`

Keep in mind that this approximation will only hold if

`(["H"_3"O"^(+)])/x xx 100% color(red)(< 5%)`

This means that you have

`x = (3.16 * 10^(-3))^2/(1.8 * 10^(-5)) = 0.55`

Make sure that the approximation holds

`(3.16 * 10^(-3) color(red)(cancel(color(black)("M"))))/(0.55color(red)(cancel(color(black)("M")))) xx 100% = 0.57% color(red)(<5%)`

Therefore, you have

`color(darkgreen)(ul(color(black)("concentration CH"_3"COOH" = "0.55 M")))`

The answer is rounded to two sig figs, the number of decimal places you have for the `"pH"` of gastric juice.