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Topic: Determining the Ksp For Ca(OH)2  (Read 22131 times)

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Offline Thomas_IonFlask

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Re: Determining the Ksp For Ca(OH)2
« Reply #15 on: April 04, 2008, 04:07:18 PM »
Shouldn't my initial Initial volume of HCL(mL) for trial 1 be zero?

Offline Thomas_IonFlask

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Re: Determining the Ksp For Ca(OH)2
« Reply #16 on: April 04, 2008, 04:12:46 PM »
I mean, Shouldn't my initial Initial volume of HCL(mL) for trial 1 be 50 mL?

Offline alley45

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Re: Determining the Ksp For Ca(OH)2
« Reply #17 on: April 04, 2008, 08:28:50 PM »
hey guys..

i have a very similar assignment, and i've been stuck on this part below for over 2 hours now. Could anyone explain this to me.

calculate the solubility product for calcium hydroxide?

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 may have gotten myself side tracked in the math, but if someone else can varify, I believe that is a correct way to reach the answer.

Offline flightman233

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Re: Determining the Ksp For Ca(OH)2
« Reply #18 on: April 04, 2008, 09:11:50 PM »
nope, thomas

Before you start adding anything you are at 0.

IT could have been one if and only if that is where you read the buret as your starting point.

But this depends on you, and where you started.

Good luck

Offline ljc560

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Re: Determining the Ksp For Ca(OH)2
« Reply #19 on: April 04, 2008, 09:24:04 PM »
I have the same question:
  When I'm Finding the molarity of the OH-, shouldn't i use the formula Ma x Va = Mc x Vc, and in this case, Ma=2.636 mL and Va=0.1 M Mc=unknown but what about Vc=??. thanks

Offline Borek

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Re: Determining the Ksp For Ca(OH)2
« Reply #20 on: April 05, 2008, 02:15:04 AM »
For every 2 Mol of HCl added you neutralize one mol of CaOH2 right?

Ca(OH)2. But yes.

Quote
Take the number of mol of hydroxide neutralized and divide by the volume of solution this will give you molarity of the CaOH2. 

Ca(OH)2. And you have correctly stated in the next step, that there is 2:1 ratio of hydroxide and Ca(OH)2 molarity, so your result now is twice too high.

Quote
You know from the stoichiometry that there is 2 times as much OH- present as calcium.

OK

Quote
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 may have gotten myself side tracked in the math, but if someone else can varify, I believe that is a correct way to reach the answer.

One small mistake, otherwise OK :)
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Offline Borek

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Re: Determining the Ksp For Ca(OH)2
« Reply #21 on: April 05, 2008, 02:30:09 AM »
I have the same question:
  When I'm Finding the molarity of the OH-, shouldn't i use the formula Ma x Va = Mc x Vc, and in this case, Ma=2.636 mL and Va=0.1 M Mc=unknown but what about Vc=??. thanks

Name your symbols. What does Mc stand for?

Your formula doesn't account for stoichiometry, it works only for systems reacting 1:1.
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