Question

`"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?

Answer 1

`"pH" = - log(sqrt(0.269 * K_a))`

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 `"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.

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

By definition, the acid dissociation constant, `K_a`, is equal to

`K_a = (["CH"_3"COO"^(-)] * ["H"_ 3"O"^(+)])/(["CH"_3"COOH"])`

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` `"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` `"M"` dissociates, the initial concentration of the acid decreases by `x` `"M"`.

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

`K_a = (x * x)/(0.269 - x)`

`K_a = x^2/(0.269 - x)`

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 ~~ 0.269`

This means that you have

`K_a = x^2/0.269`

which gets you

`x = sqrt(0.269 * K_a)`

Since we've said that `x` `"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 `"pH"` of the solution is given by

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

In your case, this is equal to

`"pH" = - log(sqrt(0.269 * K_a))`

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.