What indicator is used in this experiment?

A - Phenolpthalein

B - Bromothymol Blue

C - NaOH

D - Pthalic acid

Phenolpthalein is the acid/base indicator used in this lab.

What is the chemical formula of the acid used to standardize the NaOH solution?

A - HCl

B - C8H6O4

C - H2SO4

D - HNO3

You used phthalic acid to standardize the NaOH in part A of the lab.

During standardization, what is the colour of the solution in the Erlenmeyer flask at the equivalence point?

A - White

B - Clear/Colourless

C - Blue

D - Pink

Phenolthalein will go from colourless to pink when the pH rises

Define equivalence point.

The equivalence point is where you have stoichiometric equivalent amounts of acid and base.

Write the equation for the reaction of NaOH with the indicator. (Use Hind as the chemical formula)

NaOH + HInd -> Na+Ind- + H2O.

What is the main property of NaOH that prevents us from knowing its molarity accurately without standardization?

Solid NaOH is hygroscopic, and absorbs moisture from the air, making exact weighing very difficult in normal laboratory conditions.

Due to the small screen width we could not display a graph with additional questions. Please view on a larger screen to view these extra problems.

The above is a simulation of a titration of 10.00 mL of an unknown acid with 0.1022 M NaOH. We can use it to simulate the types of results you will obtain during your lab.

You can set the volume you will titrate by changing the numerical value in the Amount of base titrated (mL) Your virtual buret can only titrate values to a precision of 0.01 mL.

You can click on titrate in order to simulate the addition of the stated volume.

You can restart the titration by pressing the clear data button

Finally, you can download your results as a text file that resembles closely the file that will be generated in lab.

Use the simulation to answer the following questions:

You are performing the titration of 10.00 mL of an unknown weak acid with 0.1022 M NaOH that you have previously prepared.

What is the intial pH of the acid before adding any base.

From the graph you can see the intial pH is ~2.4.

Titrate the acid with 1.0 mL portions of NaOH until you are past the equivalence point (~10 mL). Estimate from the graph generated the pH and volume of base added at the equivalence point.

pH = ~ 8

Volume = ~ 9.8 mL

Using such a large titration volume, it can be very difficult to determine the pH and volume at the equivalence point. This is why it is very important to reduce our titration volume as we get close to the equivalence point.

Clear the simulation. This time run it as you will the experiment. Try to titrate a volume that raises the pH by about 0.15 units each titration. As you get closer to the equlibration point, you will need to reduce the volume that you titrate in order to keep the pH from rising more than 0.15 per addition. Once you get very close to the equilibration point it will be almost impossible to keep it under 0.15, so try to use the lease volume of titrant as possible to increased the number of data points you have near the equivalence point.
What volume of base is neccesary in order to reach the equivalence point. (You may want to download the data and use the first derivitave excel file on MyCourses to help you determine the equivalence point)

9.79 mL

Determine the pKa of the unknown acid by determining the pH at the titration half point.

3.8

Find the pH at half the volume used to get to the equivalence point. At this point pH = pKa for your acid.

Determine the identity of the unknown acid by using the pKa of the acid.

A - Chloroacetic acid 2.85

B - Formic acid 3.75

C - Benzoic acid 4.19

D - Acetic acid 4.75

Formic acid

The pKa of the acid determined in Q.9 of 3.8 matches best with the pKa of formic acid.

Determine the concentration of the unknown acid.

0.100 M

The concentraion of acid can be used by using C1V1 = C2V2 and using the volume at the equivalence point.

C1 = C2V2/ V1

C1 = 0.1022 M X 9.79 mL / 10.00mL

C1 = 0.100 M

Now that you know the actual pKa of the acid from the data table, calculate the intial pH of the solution before any base has been added.

2.378

Set up our ICE table with x = [HCOOH]dissoc = [HCOO-] = [H3O+]:

Concentration (M) HCOOH(aq) + H2(l) $\ce{ <=>}$ H3O+(aq) + HCOO-(aq)
Initial 0.100 - 0 0
Change -x - +x +x
Equilibrium 0.100-x - x x

Since our Ka is small compared to [HCOOH]init we can assume [HCOOH]init-x = [HCOOH]init = 0.100 M

We can substitute our expression for Ka

$K_{a} = \frac{[\ce{H3O^{+}}][\ce{HCOO^{-}}]}{ [\ce{HCOOH}]} $

= 1.75 x 10-4

~= $\frac{x*x}{0.100 \, M}$

$ x = \sqrt{(0.100 \,M)(0.000175)} = $ 4.18 x 10-3

pKa = -log(4.18 x 10-3)

pKa = 2.378

This is very close to our previous value obtained from the graph of ~2.4