Purpose: The purpose of this experiment is to determine the Ksp for calcium hydroxide.

Question

Materials are:
*50.0 mL of 0.10 M HCl(aq)
*bromophenol blue indicator
*50.0 mLsaturated calcium hydroxide
*50.0 mL buret
*125 mL Erlynmyer Flask
*50.0 mL of distilled water
*10.0 mL pipet with pump
*waste beaker

The procedure for the lab is:
1.Fill the buret with the acid solution.
2.Pipet 10.0 mL of calcium hydroxide into the flask.
3.Add approximately 10.0 mL of water to the flask.
4.Add 4 drops of bromophenol blue to the flask.
5.Titrate the base to an endpoint and record the volume of acid added.
6.Dispose of the flask contents into a waste beaker.
7.Repeat steps 2-6 four more times.

The following is what I am having trouble with.

Present your data in a table.Use the following table as a guide:
Trial 1 Trial 2 Trial 3
*Initial volume
of HCL(mL)

*Final volume
of HCL (mL)

*Volume of
HCL added (mL)

I am having difficulty regarding the trial 1 column. What is my initial volume of HCL(mL) supposed to be?

Answer 1

Could it be the reading on the burette? As 50mL which was called for?

Answer 2

I think the instructions are not too good, if I interpret them correctly. I THINK the initial volume is the initial READING OF THE BURET. Then I think the final volume is the FINAL READING OF THE BURET. Subtract the initial reading from the final reading to obtain the volume of HCl added (in mL).
For example:
The initial reading of the buret may have been something like 0.72 mL. Then the titration is performed and the final reading at the end point of the indicator is 4.52 mL. The volume of HCl used for the titration then is 4.52 - 0.72 = ?? Just as a side note, when I do a titration, I ALWAYS make the initial reading 0.00 although teachers have told me through the years that I'm wasting time to do that. But it has some advantages, at least for me.
1. I ALWAYS know what the initial reading was since it never changes.
2. I don't need to make an entry into the notebook of the initial volume.
3. I need not make a big deal out of subtracting (a possible source of a math error) because the final reading IS the volume of titrant used.
I always advise students to follow the instructions provided by their prof. Personally, I prefer the way I do it.

Answer 3

Here are my results:

*Initial volume
of HCL(mL)
Trial 1- 0.0
Trial 2- 2.51
Trial 3- 4.95
Trial 4- 7.4
Trial 5- 9.9

*Final volume
of HCL (mL)
Trial 1- 2.51
Trial 2- 4.95
Trial 3- 7.4
Trial 4- 9.9
Trial 5- 12.39

*Volume of
HCL added (mL)
Trial 1- 2.51
Trial 2- 2.44
Trial 3- 2.45
Trial 4- 2.50
Trial 5- 2.49

But now the next question is asking me to calculate the solubility product for calcium hydroxide. How do I do this? I would be very thankful for your help

Answer 4

Ca(OH)2 ==> Ca^+2 + 2OH^-
Ksp = (Ca+2)(OH^-)^2 = ??
You know volume of HCl. You know the molarity of the HCl.
mols HCl = L x M = ??
The equation for the titration is
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O

So take 1/2 mols HCl to give you mol OH^-.
mols OH^-/volume of the sample = (OH^-) in mols/L. Plug that into Ksp expression. 1/2 of the (OH^-) is Ca^+2.

Answer 5

To get volume of HCL, would I do this?

2.51+2.44+2.45+2.50+2.49 / 5

= 2.48

?

Answer 6

No, although that would give you an average and in the end you will want to calculate an average anyway. However, I think you want to use each one as a separate experiment because you will want to see how good the precision is. That way you will have five experimental determinations of Ksp. Use 2.51 mL for the first one.

Answer 7

I see, so for the first trial my volume would be 2.51 ml

For the second trial, the volume would be 2.44 ml

For the third trial, the volume will be
2.45 mL

For the fourth trial, the volume will be 2.50 mL

For the fifth trial, the volume will be 2.49 mL

?

Answer 8

right.

Answer 9

Im going to try to solve for the first trial please let me know if I am correct. I will send it within the next 10 minutes