# Compute pH before/after addition of NaOH?

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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.

In the titration of 50.0 mL of 0.100 M

##### 1 Answer

Here's what I got.

#### Explanation:

**!! EXTREMLEY LONG ANSWER !!**

For the sake of simplicity, I'll use

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

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

If you take

#[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 = 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, **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 **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**

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 **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"-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

#[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 = 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,