# chemistry

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

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1. It looks ok to me but just to be on the safe side, I have added a little to one of your sentences..
Take the number of mol of Calcium hydroxide neutralized and divide by the volume of solution this will give you molarity of the CaOH2Ca(OH)2.

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posted by DrBob222
2. I have one last question.

Conduct a search to determine the actual value for the solubility product of calcium hydroxide. Compare your experimental value to this value by determining the percentage difference.

How would I use the percentage difference formula in this case to arrive at the answer?

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posted by Jessica
3. Here is one site that lists 5.5 x 10^-6.

(difference between values)*100/accepted value = % difference or % error.

When I do these I use the absolute value of the difference BUT if you want to show that your value is higher or lower, then difference must be [(experimental value - accepted value)/accepted value] *100

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posted by DrBob222
4. Here is the site.
http://www.csudh.edu/oliver/chemdata/data-ksp.htm

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posted by DrBob222

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