Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: sharbeldam on November 14, 2020, 05:21:17 AM
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In this lab we have a saturated solution of calcium hydroxide, to calculate the concentration of OH-, we calculate pH of the solution initially, my first question is that is that necessary? if we have Ksp, can't we just do the ksp equation to find out the concentration without calculating the pH?
Ksp=S^2 * S and then we can get S.
do we use Hcl and titration for the calculation to be experimental and not in theory?
my second question is the second part where we calculate the solubility of Calcium hydroxide in calcium chloride solution in the same way -> Titration
let's say i use 20ml of HCl 2M , i can get the moles but in the lab instruction im given these equations:
Ca(OH)2 + 2HCl ---> Ca+2 + 2OH- + H2O
Ca(OH)+ + HCl --> Ca+ +2OH- +H2O
this second reaction that also happens according to the professor is what confused me, so let's say i get moles of HCl added, how do i figure it the concentration of hydroxide and hence the solubility of Ca(OH)2.
I'd be happy for some hints, thank you
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Ca(OH)2 + 2HCl ---> Ca+2 + 2OH- + H2O
Ca(OH)+ + HCl --> Ca+ +2OH- +H2O
Check material balance for both reactions.
As fae as I know Ca1+ is not known yet.
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I truly apologize
it was
Ca(OH)+ + HCl --> Ca+2 + OH- + H2O
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Base + acid is still a base? Where are the chloride ions?
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Im so stressed because of understanding this lab so i cant type so here is the image.
but my problem is the calculations, since there are 2 equations not one...
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I am afraid so far you have failed to explain what is the lab, what you are expected to do, and what is the problem you have. Everything you posted is way too chaotic.
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You are right, I did mess up, but i think the first part of my question is clear, the second part is that i have to find out the concentration of OH- (aka solubility) of Ca(OH)2 solution in CaCl2 (common ion is Ca)
i titrate it with HCl to find out that, but the problem is that im given two equation (i posted in previous post) so how can i calculate the concentration, its not just a simple acid-base 1 equation but two...
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Your second equation is incorrect. It should be:
CaOH+ + OH- + 2HCl =... .
Therefore, from the titration you calculate the concentration of OH- ions, then calcium as half of the neutralized OH- ions.
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That's not my equation, its the college notes, they said these two reactions happen haha.
but ok, could be mistake so to calculate
Can I just use the first equation?
Ca(OH)2 + 2HCl ---> 2H2O + CaCl2
if i titrated let's say 5 moles of the acid, then i would know i have 5 moles of OH- and then i can find the concentration of the hydroxide ion and Ca+2 will be half of that. which brings me back to my initial questions.
Why do i need to titrate it to find out that?
if i know that this solution is saturated and i have the ksp, cant i just use the ksp equation?
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Do you want to check how the addition of calcium ions affects the solubility of calcium hydroxide?
Then you have to do the titration of the calcium hydroxide itself and (probably) a few clear solutions (supernatant) with the addition of calcium salts.
You can compare the result of the titration of calcium hydroxide alone with the value calculated from the solubility product. If you do the titration properly, you can find out that you have more calcium hydroxide in the solution than it results from the solubility product because the solubility product does not include the CaOH+ ion.
The second equation simply shows that you understand the phenomena in the solution more. and possibly useful for discussing exercises.
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So if i understand correctly, when I titrate the solution of Ca(OH)2+CaCl2 (With extra Ca+2 ions) I will get lower [OH-] concentration.
In the first solution to calculate the concentration of calcium , i can just divide the concentration of [OH-] by 2.
but what about solution number 2, I can't do the same right? will I have to do the Ice table to calculate the concentration of Ca+2?
Ca(OH)2 <---> Ca+2 + 2OH-
0.1m 0
+s +2s
...etc?
I appreciate your patience and support, thank you
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What the second solution?
I don't have instruction for your lab exercise, you can't make the problem clear because you don't understand it. For each laboratory exercise, there is an instruction and recommended literature - usually 30-50 pages. You have to read it several times. Do it, fill out the form, and if you don't understand something, post it on the forum along with the manual and ask meaningful questions.
On many occasions in this Forum, there are cases where it takes days to fix a problem that can be solved in minutes or less.
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By second solution I meant the solution of Ca(OH)2+CaCl2 (that tests for the common ion effect)
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Here we have a problem. Qualitatively, turbidity can be observed. I will not speak further because I do not know the details of the procedure.
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Ok, i appreciate your help ty