A #"25.0-mL"# sample of #"0.150-mol L"^(-1)# acetic acid is titrated with a #"0.150-mol L"^(-1)# #"NaOH"# solution. What is the #"pH"# at the equivalence point?

!! LONG ANSWER !!

The trick here is to realize that in order to reach the equivalence point, you need to add enough moles of sodium hydroxide, which I'll represent as hydroxide anions because this compound is a strong base, to consume all the moles of acetic acid present in the initial solution.

For a weak acid - strong base titration, you have

#color(blue)(ul(color(black)("equivalence point "))) = {("all the moles of the weak acid are consumed"), ("all the moles of the strong base added are consumed") :}#

As you know, acetic acid and sodium hydroxide neutralize each other in a #1:1# mole ratio to produce acetate anions, the conjugate base of the weak acid, and water.

#"CH"_ 3"COOH"_ ((aq)) + "OH"_ ((aq))^(-) -> "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#

Now, use the molarity and the volume of the acetic acid solution to determine how many moles of acetic acid are present in the initial solution.

#25.0 color(red)(cancel(color(black)("mL solution"))) * ("0.150 moles CH"_ 3"COOH")/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.00375 moles CH"_3"COOH"#

This means that in order to reach the equivalence point, you need to add #0.00375# moles of sodium hydroxide.

Since the sodium hydroxide solution has the same molarity as the acetic acid solution, it follows that you will need to add #"25.0 mL"# of the sodium hydroxide solution to deliver #0.00375# moles of hydroxide anions to the reaction.

This means that after you mix the two solutions, the total volume of the resulting solution will be

#"25.0 mL + 25.0 mL = 50.0 mL"#

Now, notice that for every mole of acetic acid and of hydroxide anions that the reaction consumes, you get #1# mole of acetate anions.

Since the reaction consumes #0.00375# moles of acetic acid and of hydroxide anions, you can say that after the reaction is complete, the resulting solution will contain #0# moles of acetic acid and of hydroxide anions and #0.00375# moles of acetate anions.

The concentration of the acetate anions in the resulting solution will be--do not forget to convert the volume of the solution to liters!

#["CH"_ 3"COO"^(-)] = "0.00375 moles"/(50.0 * 10^(-3) quad "L") = "0.0750 mol L"^(-1)#

Now, the acetate anions will act as a weak base in aqueous solution. These anions will react with water to reform some of the weak acid and produce hydroxide anions.

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

At this point, it should become clear that the #"pH"# of the resulting solution will be #>7# because of the presence of the hydroxide anions.

As you know, an aqueous solution at #25^@"C"# has

#color(blue)(ul(color(black)(K_a * K_b = 1.0 * 10^(-14))))#

Here

• #K_a# is the acid dissociation constant of the weak acid
• #K_b# is the base dissociation constant of the conjugate base

In your case, you have

#K_b = (1.0 * 10^(-14))/(1.8 * 10^(-5)) = 5.556 * 10^(-10)#

By definition, the base dissociation constant for the ionization of the acetate anions is equal to

#K_b = (["CH"_ 3"COOH"] * ["OH"^(-)])/(["CH"_3"COO"^(-)])#

If you start with #"0.0750 mol L"^(-1)# of acetate anions and say that the reaction will consume #x# #"mol L"^(-1)# of acetate anions, you can say that, at equilibrium, the resulting solution will contain

#["CH"_ 3"COOH" ] = ["OH"^(-)] = x quad "mol L"^(-1)#

For every mole of acetate anions that reacts with water, you get #1# mole of acetic acid and #1# mole of hydroxide anions.

#["CH"_3"COO"^(-)] = (0.075 - x) quad "mol L"^(-1)#

For every mole of acetate anions that reacts with water, the number of moles of acetate anions decreases by #1# mole.

Plug this into the expression of the base dissociation constant to get

#K_b = (x * x)/(0.075 - x)#

#K_b = x^2/(0.075 - x)#

Now, because the value of base dissociation constant is significantly smaller than the initial concentration of the acetate anions, you can use the approximation

#0.075 - x ~~ 0.075#

This means that you have

#K_b = x^2/0.075#

which gets you

#x = sqrt(0.075 * 5.556 * 10^(-10)) = 6.455 * 10^(-6)#

Since #x# represents the equilibrium concentration of the hydroxide anions, you can say that the resulting solution has

#["OH"^(-)] = 6.455 8 10^(-6) quad "mol L"^(-1)#

Finally, you know that an aqueous solution at #25^@"C"# has

#color(blue)(ul(color(black)("pH + pOH = 14")))#

Since

#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))#

you can say that the #"pH"# of the solution is equal to

#"pH" = 14 + log(["OH"^(-)])#

Plug in your value to find

#"pH" = 14 + log(6.455 * 10^(-6)) = color(darkgreen)(ul(color(black)(8.81)))#

I'll leave the answer rounded to two decimal places, but keep in mind that you have three sig figs for your values, so you should report the answer as

#"pH" = 8.810#

Notice that the #"pH"# of the resulting solution is indeed #>7#. This happens because when you reach the equivalence point in a weak acid - strong base titration, the solution will contain the conjugate base of the weak acid, which will act as a weak base and react with water to produce hydroxide anions.