"20.0 mL" of glacial acetic acid (pure "CH"_3"COOH") is diluted to "1.30 L" with water. The density of glacial acetic acid is "1.05 g/mL". What is the pH of the resulting solution?

Apr 3, 2018

$\text{pH} = - \log \left(\sqrt{0.269 \cdot {K}_{a}}\right)$

Explanation:

Start by using the density of glacial acetic acid to calculate the mass of acetic acid present in the solution.

20.0 color(red)(cancel(color(black)("mL"))) * "1.05 g"/(1color(red)(cancel(color(black)("mL")))) = "21.0 g"

Next, use the molar mass of acetic acid to calculate the number of moles present in the solution.

21.0 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COOH")/(60.05color(red)(cancel(color(black)("g")))) = "0.3497 moles CH"_3"COOH"

The total volume of the solution is equal to $\text{1.30 L}$, so use it to find the molarity of the acid.

["CH"_3"COOH"] = "0.3497 moles"/"1.30 L" = "0.269 M"

Now, acetic acid is a weak acid, which implies that it will only partially ionize in aqueous solution.

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

By definition, the acid dissociation constant, ${K}_{a}$, is equal to

${K}_{a} = \left(\left[\text{CH"_3"COO"^(-)] * ["H"_ 3"O"^(+)])/(["CH"_3"COOH}\right]\right)$

Notice that every mole of acetic acid that dissociates produces $1$ mole of acetate anions and $1$ mole of hydronium cations. This means that if you take $x$ $\text{M}$ to be the concentration of acetic acid that dissociates, you can say that, at equilibrium, the solution will contain

["CH"_ 3"COO"^(-)] = ["H"_ 3"O"^(+)] = x quad "M"

Similarly, the equilibrium concentration of the acetic acid will be

["CH"_3"COOH"] = (0.269 - x) quad "M"

This means that when $x$ $\text{M}$ dissociates, the initial concentration of the acid decreases by $x$ $\text{M}$.

Pug this into the expression of the acid dissociation constant to get

${K}_{a} = \frac{x \cdot x}{0.269 - x}$

${K}_{a} = {x}^{2} / \left(0.269 - x\right)$

Now, you didn't provide a value for the acid dissociation constant, but I know from experience that this value is small enough to allow you to use the following approximation.

$0.269 - x \approx 0.269$

This means that you have

${K}_{a} = {x}^{2} / 0.269$

which gets you

$x = \sqrt{0.269 \cdot {K}_{a}}$

Since we've said that $x$ $\text{M}$ represents the equilibrium concentration of the hydronium cations (and of the acetate anions), you will have

["H"_ 3"O"^(+)] = sqrt(0.269 * K_a) quad "M"

Finally, the $\text{pH}$ of the solution is given by

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

In your case, this is equal to

$\text{pH} = - \log \left(\sqrt{0.269 \cdot {K}_{a}}\right)$

Now all you have to do is to use the value of the acid dissociation constant given to you--or do a quick search for the acid dissociation constant of acetic acid--and plug it into the equation.

The value should be rounded to three decimal places because you have three sig figs for your values.