Compute pH before/after addition of NaOH?

In the titration of 50.0 mL of 0.100 M #\beta#-hydroxybutyric acid, #HC_4H_7O_3#, with 0.100 M NaOH, compute pH before addition of NaOH, and after the addition of 25.00 mL and 50.00 mL of NaOH. #pK_a# for #HC_4H_7O_3# is 4.39.

1 Answer
Mar 29, 2017

Answer:

Here's what I got.

Explanation:

!! EXTREMLEY LONG ANSWER !!

For the sake of simplicity, I'll use #beta#-#"HA"# to denote the acid and #beta#-#"A"^(-)# to denote the conjugate base.

Now, the pH of the solution before the addition of strong base can be calculated by using the fact that the weak acid only partially ionizes in a #1:1# mole ratio to produce hydronium cations and #beta#-hydroxybutyrate anions

#beta"-HA"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + beta"-A"_ ((aq))^(-)#

If you take #x# to be the equilibrium concentration of the hydronium cations and of the #beta#-hydroxybutyrate anions, you can say that the equilibrium concentration of the acid will be

#[beta"-HA"] = [beta"-HA"]_0 - x#

By definition, the acid dissociation constant will be

#K_a = ( [beta"-A"^(-)] * ["H"_3"O"^(+)])/([beta"-HA"])#

In your case, this will be equal to

#10^(-"p"K_a) = (x * x)/(0.100 - x) = x^2/(0.100 - x)#

Since

#K_a = 10^(-"p"K_a) = 10^(-4.39) = 4.07 * 10^(-5)#

is very small compared to the initial concentration of the acid, you can use the approximation

#0.100 -x ~~ 0.100#

This means that you have

#4.07 * 10^(-5) = x^2/0.100#

Solve for #x# to get

#x = sqrt(0.100 * 4.07 * 10^(-5)) = 2.02 * 10^(-3)#

You can thus say that

#["H"_3"O"^(+)] = 2.02 * 10^(-3)# #"M"#

The pH of the initial solution

#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)])))#

will be

#"pH" = - log(2.02 * 10^(-3)) = color(darkgreen)(ul(color(black)(2.695)))#

Now, #beta#-hydroxybutyric acid will react with the hydroxide anions provided by the strong base in a #1:1# mole ratio to produce #beta#-hydroxybutyrate, its conjugate base, and water.

#beta"-HA"_ ((aq)) + "OH"_ ((aq))^(-) -> beta"-A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#

The initial solution contains

#50.0 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.100 moles" color(white)(.)beta"-HA")/(1color(red)(cancel(color(black)("L")))) = "0.00500 moles"# #beta"-HA"#

The first solution of sodium hydroxide contains

#25.00 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.100 moles OH"^(-))/(1color(red)(cancel(color(black)("L")))) = "0.00250 moles OH"^(-)#

So, after the first sample of sodium hydroxide is added to the initial solution, the moles of hydroxide anions will be completely consumed. You will be left with

#n_ ("OH"^(-)) = 0 -># completely consumed

#n_ (beta"-HA") = "0.00500 moles" - "0.00250 moles" = "0.00250 moles"# #beta"-HA"#

#n_ (beta"-A"^(-)) = "0 moles" + "0.00250 moles" = "0.00250 moles"# #beta"-A"^(-)#

The new volume of the solution will be

#V_"solution" = "50.0 mL" + "25.00 mL" = "75.0 mL"#

Now, notice that the resulting solution contains equal numbers of moles of weak acid and of conjugate base. This means that you're in the buffer region, i.e. the resulting solution contains a buffer.

This means that the pH of the solution will be equal to the #"p"K_a# of the weak acid because the concentrations of the weak acid and of the conjugate base are equal.

#[beta"-HA"] = [beta"-A"^(-)]#

This is known as the half-equivalence point. Consequently, you will have

#"pH" = "p"K_a + log ( (color(red)(cancel(color(black)([beta"-A"^(-)]))))/(color(red)(cancel(color(Black)([beta"-HA"]))))) -># the Henderson - Hasselbalch equation

and so

#color(darkgreen)(ul(color(black)("pH" = 4.39)))#

Finally, you're adding an additional #"25.00 mL"# of strong base to get the total volume of strong base added to #"50.00 mL"#.

At this point, it should be obvious that all the moles of acid and all the moles of hydroxide anions will be consumed by the reaction #-># this is known as the equivalence point of the titration. You will be left with

#n_ ("OH"^(-)) = 0 -># completely consumed

#n_ (beta"-HA") = "0.00250 moles" - "0.00250 moles" = 0 -># completely consumed

#n_ (beta"-A"^(-)) = "0.00250 moles" + "0.00250 moles" = "0.00500 moles"# #beta"-A"^(-)#

The new volume of the solution will be

#V_"total" = "75.0 mL" + "25.00 mL" = "100.0 mL"#

The concentration of the conjugate base will be equal to

#[beta"-A"^(-)] = "0.00500 moles"/(100.0 * 10^(-3)"mL") = "0.0500 M"#

The conjugate base will only partially ionize to produce #beta#-hydroxybutyric acid and hydroxide anions in #1:1# mole ratios

#beta"-A"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons beta"-HA"_ ((aq)) + "OH" _((aq))^(-)#

An aqueous solution at room temperature has

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

and so

#K_b = 10^(-14)/(4.07 * 10^(-5)) = 2.46 * 10^(-10)#

If you take #x# to be the equilibrium concentrations of #beta#-hydroxybutyric acid and hydroxide anions, you will have

#[beta"-A"^(-)] = 0.0500 - x#

By definition, you know that

#K_b = ([beta"-HA"] * ["OH"^(-)])/([beta"-A"^(-)])#

which, in your case, is

#K_b = (x * x)/(0.0500 - x) = x^2/(0.0500 - x)#

Once again, use the approximation

#0.0500 - x ~~ 0.0500#

to get

#2.46 * 10^(-10) = x^2/0.0500#

Solve for #x# to find

#x = sqrt(0.0500 * 2.46 * 10^(-10)) = 3.51 * 10^(-6)#

This means that the resulting solution has

#["OH"^(-)] = 3.51 * 10^(-6)# #"M"#

Consequently, you will have

#["H"_3"O"^(+)] = 10^(-14)/(3.51 * 10^(-6)) = 2.85 * 10^(-9)# #"M"#

which means that the pH of the solution is

#"pH" = - log(2.85 * 10^(-9)) = color(darkgreen)(ul(color(black)(8.545)))#

I'll leave the values rounded to three decimal palces, the number of sig figs you have for the concentrations of the two solutions.

Now, does the result make sense?

You are titrating a weak acid with a strong base, so at the equivalence point, the solution will only contain the conjugate base of the acid.

So even without doing any calculations, you should be able to say that at equivalence point, #"pH" > 7#.