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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Thomas_IonFlask on April 01, 2008, 03:40:00 PM

Title: Determining the Ksp For Ca(OH)2
Post by: Thomas_IonFlask on April 01, 2008, 03:40:00 PM
Can someone please tell me what needs to be put into the Trial 1 column of the table?This is found in the ANalysis section of the lab. I have had great difficulty understanding this.

TheLab:  Determining the Ksp For Ca(OH)2

Purpose:  The purpose of this experiment is to determine the Ksp for calcium hydroxide.  

Experimental Design:  A saturated solution of calcium hydroxide is titrated with a hydrochloric acid solution of known concentration to a bromophenol blue endpoint (yellow).

Materials:

50.0 mL of 0.10 M HCl(aq)         bromophenol blue indicator
50.0 mLsaturated calcium hydroxide                50.0 mL buret               
125 mL Erlenmeyer flask         50.0 mL of distilled water
10 mL pipet with pump         waste beaker

Procedure:

1.   Fill the buret with the acid solution.
2.   Pipet 10.0 mL of calcium hydroxide into the flask.
3.   Add approximately 10.0 mL of water to the flask.
4.   Add 4 drops of bromophenol blue to the flask.
5.   Titrate the base to an endpoint and record the volume of acid added.
6.   Dispose of the flask contents into a waste beaker.
7.   Repeat steps 2-6 four more times.

Analysis:

1.   Present your data in a table.  Use the following table as a guide:

                           Trial 1   Trial 2   Trial 3   Trial 4   Trial 5
Initial volume of HCl (mL)               
Final volume of HCl (mL)               
Volume of HCl added (mL)               
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Borek on April 01, 2008, 04:55:23 PM
I don't understand what you don't understand ???

When titrating you will be filling your buret with HCl solution. You read how much acid is in the buret and write it down, then you titrate to the endpoint and you read volume and write it down again, then you subtract second from the first to find volume added.
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Thomas_IonFlask on April 01, 2008, 07:21:24 PM
SO would my Initial volume of HCl (mL) be 50 mL?
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Thomas_IonFlask on April 01, 2008, 09:23:24 PM
Let me repost the important information only:

Purpose: The purpose of this experiment is to determine the Ksp for calcium hydroxide.

Materials are:
*50.0 mL of 0.10 M HCl(aq)
*bromophenol blue indicator
*50.0 mLsaturated calcium hydroxide
*50.0 mL buret
*125 mL Erlynmyer Flask
*50.0 mL of distilled water
*10.0 mL pipet with pump
*waste beaker

The procedure for the lab is:
1.Fill the buret with the acid solution.
2.Pipet 10.0 mL of calcium hydroxide into the flask.
3.Add approximately 10.0 mL of water to the flask.
4.Add 4 drops of bromophenol blue to the flask.
5.Titrate the base to an endpoint and record the volume of acid added.
6.Dispose of the flask contents into a waste beaker.
7.Repeat steps 2-6 four more times.

The following is what I am having trouble with.

Present your data in a table.Use the following table as a guide:
                                                         Trial 1 Trial 2 Trial 3
*Initial volume
of HCL(mL)

*Final volume
of HCL (mL)

*Volume of
HCL added (mL)

I am having difficulty regarding the trial 1 column. What is my initial volume of HCL(mL) supposed to be?
Title: Re: Determining the Ksp For Ca(OH)2
Post by: flightman233 on April 01, 2008, 11:05:17 PM
As Borek stated Thomas,

When you are looking at your buret, and you fill it to the top (Oml mark) and begin adding it to the solution, you stop once you have reached the yellow color in solution.

Initially you were at 0ml of HCl added.

Finally you were at X(whatever) number of ml added. 

When you subtract the final - initial = net amount added (simply looking at the buret would tell you how much you added though if you started at 0ml)

Then apparently you had to do this 5 times, each with the same calculation as above.
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Thomas_IonFlask on April 01, 2008, 11:22:45 PM
 Initially you were at 0ml of HCl added. So, in other words, Initial volume of HCL(mL) for trial 1 is 0 mL?
Title: Re: Determining the Ksp For Ca(OH)2
Post by: flightman233 on April 01, 2008, 11:28:26 PM
Yes, if you filled the buret to the 0 ml mark, then initially at trial 1 you were at 0 ml.

Now assuming you didn't refill it on trial 2 and say you were at 12 ml.

You record 12 ml as your initial volume for that trial.

Make sense?
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Thomas_IonFlask on April 02, 2008, 12:39:39 AM
Here are my results:


*Initial volume
of HCL(mL)
Trial 1- 0.0
Trial 2- 2.51
Trial 3- 4.95
Trial 4- 7.4
Trial 5- 9.9

*Final volume
of HCL (mL)
Trial 1- 2.51
Trial 2- 4.95
Trial 3- 7.4
Trial 4- 9.9
Trial 5- ?

*Volume of
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?
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Borek on April 02, 2008, 03:47:02 AM
If you have not noted final volume after 5th titration, you will not have volume of HCl used.
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Padfoot on April 02, 2008, 06:58:39 AM
(2.51+2.44+2.45+2.5)/4  ;)
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Thomas_IonFlask on April 02, 2008, 07:34:45 AM
When I have the table filled out, what would I need to do in order to calculate the solubility product for calcium hydroxide?
Title: Re: Determining the Ksp For Ca(OH)2
Post by: flightman233 on April 02, 2008, 01:29:03 PM
Well first what does Ksp mean?

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.
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Borek on April 02, 2008, 02:10:30 PM
Ca(OH)2
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Thomas_IonFlask on April 02, 2008, 08:40:57 PM
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?
Title: Re: Determining the Ksp For Ca(OH)2
Post by: flightman233 on April 03, 2008, 08:24:33 AM


Your precent difference formula can be found in your text.

Once you know the general formula all you do is plug the value you calculated in, and then the actual value in the appropriate places.

Just a simple plug and play equation with two variables.

Good luck.


...Borek.... my mistake about the Ca(OH)2  was in kind of a rush to get the post typed   :D   but thanks for the correction!
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Thomas_IonFlask on April 04, 2008, 04:07:18 PM
Shouldn't my initial Initial volume of HCL(mL) for trial 1 be zero?
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Thomas_IonFlask 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?
Title: Re: Determining the Ksp For Ca(OH)2
Post by: alley45 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.
Title: Re: Determining the Ksp For Ca(OH)2
Post by: flightman233 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
Title: Re: Determining the Ksp For Ca(OH)2
Post by: ljc560 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
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Borek 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 :)
Title: Re: Determining the Ksp For Ca(OH)2
Post by: Borek 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.