# #"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?

##### 1 Answer

#### Answer:

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

Notice that **every mole** of acetic acid **that dissociates** produces **mole** of acetate anions and **mole** of hydronium cations. This means that if you take **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 aciddecreasesby#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

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

Finally, the

#"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.