Regarding the table, trial 1, trial 2 and trial 3 are not so neatly written but they should be shifted a little to the right
I answered this at the original post. If you still have questions, repost here.
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- ?
HCL added (mL)
Trial 1- 2.51
Trial 2- 2.44
Trial 3- 2.45
Trial 4- 2.50
Trial 5- ?
How would I get Final volume of HC (mL) and Volume of HCL added (mL) for trial 5?
You must not have read the final volume for trial 5. Either that or you have misaligned the other readings.
When I have the table filled out, what would I need to do in order to calculate the solubility product for calcium hydroxide?
The following is what I have in mind, please tell if I am incorrect or correct.
Ksp = [Ca+2][OH-1]2
You should have an average value for the amount of H+ that you put into solution.
For every 2 Mol of HCl added you neutralize one mol of CaOH2 right?
Take the number of mol of hydroxide neutralized and divide by the volume of solution this will give you molarity of the CaOH2.
You know from the stoichiometry that there is 2 times as much OH- present as calcium.
So.. plug these numbers into the equation for the Ksp and you should arrive at your answer.
Ksp = [molarity of OH- / 2 ][molarity of OH-]2
I don't like to intrude but if I may make a comment on this.
I have to say that just like the previous trials didn't you record how much you added for trial 5?
(I'm not understanding how you don't know the volume you added for trial 5)
You CAN however, theoretically calculate about how much HCl you should have added to trial 5)
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