Calculating pH Change in C6H5COOH Solution with KOH Addition

[lizzyb]

Question:

Calculate the change in pH of a 0.400 M C6H5COOH solution when 100.0 ml of a 2.50 M KOH solution is added to 1.000 L of C6H5COOH. For C6H5COOH, Ka = 6.46 X 10^-5.

Work Done So Far:

We're to determine the change in pH, thus we'll need to find the original pH.

Ka = 6.46 X 10^-5 = x^2/(.4-x) ==> x = .0050511 ==> pH = 2.296

What should I do next? I understand that KOH is a strong base so it will tear the excess H off of C6H5COOH leaving C6H5COO-, but do we need to account for the H+ left over from the C6H5COOH there before the KOH was added?

 

[symbolipoint]

Can you now use the given information to find the formality of C6H5COOH and the formality of its potassium salt? Also, is there excess KOH?

 

[lizzyb]

One idea was to count the number of moles of each and then recognize that the OH will pull the H off of C6H5COOH, so after it does that, then there will be excess OH-.

Moles Acetic: .400 M * 1L = .4 moles
Moles KOH: .250 M * .1 L = .025 moles

In this case, there is not an excess number of KOH, so all the OH is used up removing H from the acetic acid.

Moles Acetic Acid left: .4 - .025 = .375 moles

Its new molarity is: .375 moles / 1.1L = .341 M

How do I determine the pH from this? Should I just do it like I did before taking the molarity as .341 M?

 

[symbolipoint]

$$
\eqalign{
& \text{Refer to reaction} \cr
& \text{HAc } \Leftrightarrow \text{ H + + Ac - } \cr
& \text{Each mole of HAc which dissociates yields one mole of H and one mole of Ac}\text{.} \cr
& \text{Molarity of any Ac } = \;Fsalt\; + \;H \cr
& \text{Molarity of any HAc } = \;Facid\; - \;H \cr
& \text{Note that the word salt means KAc (potassium acetate)}\text{.} \cr
& \cr
& \text{Formality of remaining acid after neutralization: } \cr
& \frac{{1L \times 0.4M}}
{{1.100L}}\; - \;\frac{{0.1L \times 2.5M}}
{{1.100L}}\; = \;0.136F\,HAc \cr
& \cr
& \text{Formality of the salt KAc formed in solution is:} \cr
& \frac{{0.1L \times 2.5M}}
{{1.100L}}\; = \;0.227F\,KAc \cr
& \cr
& \text{The equilibrium constant expression for which you already have the value of }6.46 \times 10^{ - 5} \cr
& K\; = \;\frac{{[H][Fsalt + \,H]}}
{{[Facid\; - \;H]}} \cr
& \cr
& \text{Now, just put the equation into general form for a quadratic equation and find } \cr
& \text{for [H], and then simply substitute the known values and compute}\text{. You can do } \cr
& \text{similarly for the initial acid solution before any KOH were added to find the} \cr
& \text{initial [H] value}\text{.} \cr}
$$

 

[symbolipoint]

$$<br /> \eqalign{<br /> &amp; {\rm Refer to reaction} \cr <br /> &amp; {\rm HAc } \Leftrightarrow {\rm H + + Ac - } \cr <br /> &amp; {\rm Each mole of HAc which dissociates yields one mole of H and one mole of Ac}{\rm .} \cr <br /> &amp; {\rm Molarity of any Ac } = \;Fsalt\; + \;H \cr <br /> &amp; {\rm Molarity of any HAc } = \;Facid\; - \;H \cr <br /> &amp; {\rm Note that the word salt means KAc (potassium acetate)}{\rm .} \cr <br /> &amp; \cr <br /> &amp; {\rm Formality of remaining acid after neutralization: } \cr <br /> &amp; {{1L \times 0.4M} \over {1.100L}}\; - \;{{0.1L \times 2.5M} \over {1.100L}}\; = \;0.136F\,HAc \cr <br /> &amp; \cr <br /> &amp; {\rm Formality of the salt KAc formed in solution is:} \cr <br /> &amp; {{0.1L \times 2.5M} \over {1.100L}}\; = \;0.227F\,KAc \cr <br /> &amp; \cr <br /> &amp; {\rm The equilibrium constant expression for which you already have the value of }6.46 \times 10^{ - 5} \cr <br /> &amp; K\; = \;{{[H][Fsalt + \,H]} \over {[Facid\; - \;H]}} \cr <br /> &amp; \cr <br /> &amp; {\rm Now, just put the equation into general form for a quadratic equation and find } \cr <br /> &amp; {\rm for [H], and then simply substitute the known values and compute}{\rm . You can do } \cr <br /> &amp; {\rm similarly for the initial acid solution before any KOH were added to find the} \cr <br /> &amp; {\rm initial [H] value}{\rm .} \cr} <br /> $$

 

[symbolipoint]

I have been trying but the typesetting through TexAide does not work in the message.

 

[lizzyb]

did it in ascii-ese

Refer to reaction
HAc <--> H+ + Ac-

Each mole of HAc which dissociates yields one mole
of H and one mole of Ac

Molarity of any Ac = Fsalt + H
Molarity of any HAc= Facid - H

Note that the word salt means KAc (potassium acetate).

Formality of remaining acid after neutralization:

   1L * .04M     .1L * 2.5M
 ------------ - ---------- = .136F HAc
   1.100 L        1.100 L

Formality of the salt KAc formed in solution is:

   .1L * 2.5M
   ----------- = .227 F KAc
     1.100 L

The equilibrium constant expression for which you already have
the value of 6.46 X  10^-5

    [H][Fsalt + H]
K = ---------------
     [Facid - H]

Now, just put the equation into general form for a quadratic equation and find
for [H], and then simply substitute the known values and compute. You can do
similarly for the initial acid solution before any KOH were added to find the
initial [H] value

 

[lizzyb]

there is the formula

                [salt]
pH = pKa + log(--------)
                [acid]

             2.5M * .1 L
so [salt] = ------------- = .227 M
              1.1 L

              .4 M * 1 L
    [acid] = --------------- = .36 M
                1.1 L
                       [salt]
pH = -log(Ka) + log( ---------) = .399
                       [acid]

is that okay? thanks for your help.

 

[symbolipoint]

One more try, copied from Jarte wordprocessor:

It WILL NOT paste.

You appear to have most of the details well in place, but a couple of the calculated values are different than mine. Probably no fundamental big deal.

 

[Borek]

Since when benzoic acid is acetic acid?

This is classic buffer question.

You start with 0.4 mole of weak acid.

You add 0.25 mole of strong base.

After neutralization you have 0.15 mole of weak acid left and 0.25 mole of conjugated base present in the solution.

You put these values into the Henderson-Hasselbalch equation. You even don't have to calculate concentrations, as the volume cancels out, so numbers of moles are enough.

If you really want to, you may check the calculations using (free trial will do).