Could it be the reading on the burette? As 50mL which was called for?
Here are my results:
Trial 1- 0.0
Trial 2- 2.51
Trial 3- 4.95
Trial 4- 7.4
Trial 5- 9.9
of HCL (mL)
Trial 1- 2.51
Trial 2- 4.95
Trial 3- 7.4
Trial 4- 9.9
Trial 5- 12.39
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
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).
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.
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.
To get volume of HCL, would I do this?
2.51+2.44+2.45+2.50+2.49 / 5
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.
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
For the fourth trial, the volume will be 2.50 mL
For the fifth trial, the volume will be 2.49 mL
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
chemistry - Hi Cna someone please tell me what I must out into the Trial 1 ...
chemistry - Let me repost the important information only: Purpose: The purpose ...
Chemistry - Ksp - i did an experiment where we titrated HCl on calcium ...
chemistry - Determine the number of moles of calcium per mole of calcium ...
chemistry - A 0.500 L solution of 7.50 M hydrochloric acid is used to neutralize...
Chemistry - a solution saturated in calcium hydroxide(limewater) has a pH of 12....
Chemistry - 1.Write a mathematical equation that can be used to calculate the ...
chemistry - Calcium reacts with water to yield calcium hydroxide and water ...
chemistry - calcium hydroxide reacts in a neutralization reaction with sulfuric ...
chemistry - When I have the table filled out, what would I need to do in order ...