# Acetic acid + koh

1. Apr 14, 2007

### 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?

2. Apr 14, 2007

### 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?

3. Apr 14, 2007

### 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?

Last edited: Apr 14, 2007
4. Apr 14, 2007

### 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}

5. Apr 14, 2007

### symbolipoint

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

Last edited: Apr 14, 2007
6. Apr 14, 2007

### symbolipoint

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

7. Apr 14, 2007

### lizzyb

did it in ascii-ese
Code (Text):

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

8. Apr 14, 2007

### lizzyb

there is the formula

Code (Text):

[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.

9. Apr 14, 2007

### 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.

10. Apr 16, 2007

### Staff: Mentor

Since when benzoic acid is acetic acid?

This is classic buffer question.