# Question 6b487

May 14, 2017

$\text{0.55 M}$

#### Explanation:

Start by calculating the concentration of hydronium cations needed to produce a $\text{pH}$ equal to that of gastric juice.

You know that

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

so

$\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$

$\left[{\text{H"_3"O}}^{+}\right] = {10}^{- 2.50} = 3.16 \cdot {10}^{- 3}$ $\text{M}$

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

${\text{CH"_ 3"COOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

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

$\left[{\text{CH"_3"COO"^(-)] = ["H"_3"O}}^{+}\right]$

If you take $x$ $\text{M}$ to be the initial concentration of the acid, you can say that, at equilibrium, you will have

$\left[{\text{CH"_3"COOH"] = x - ["H"_3"O}}^{+}\right]$

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

${K}_{a} = \left(\left[\text{CH"_3"COO"^(-)] * ["H"_3"O"^(+)])/(["CH"_3"COOH}\right]\right)$

$1.8 \cdot {10}^{- 5} = \frac{3.16 \cdot {10}^{- 3} \cdot 3.16 \cdot {10}^{- 3}}{x - 3.16 \cdot {10}^{- 3}}$

Now, because you have such a small value for ${K}_{a}$, you can assume that

$x - 3.16 \cdot {10}^{- 3} \approx x$

Keep in mind that this approximation will only hold if

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

This means that you have

$x = {\left(3.16 \cdot {10}^{- 3}\right)}^{2} / \left(1.8 \cdot {10}^{- 5}\right) = 0.55$

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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{concentration CH"_3"COOH" = "0.55 M}}}}$

The answer is rounded to two sig figs, the number of decimal places you have for the $\text{pH}$ of gastric juice.