# Question #6b487

##### 1 Answer

#### Explanation:

Start by calculating the concentration of hydronium cations needed to produce a

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 **mole** of acetate anions and **mole** of hydronium cations.

This means that, at equilibrium, you have

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

If you take

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

The concentration of the aciddecreasesby the concentration that ionizes.

By definition, the acid dissociation constant,

#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

#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