Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: KhemDunce on April 10, 2007, 09:38:04 PM
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Hey Chem geniuses!
I am a struggling high school chem student who couldn't combine water with water without blowing something up so some help would definitley be appreciated.
My chem final has been issued already and it is as follows:
Put 5 mL of 3 M aluminum chloride solution in each of two test tubes. To one test tube add a few drops of 3 M sodium hydroxide solution. Then add more sodium hydroxide solution, with stirring, until the precipitate dissolves. Treat the other test tube similarly, but use 3 M ammonum hydroxide solution to obtain precipitation and then add more ammonium hydroxide solution. Why did the precipitate dissolve in one case and not the other?
My first problem is finding how much of each solution I need in grams. The "3 M" of X solution is flummoxing me a bit. (We haven't learned it and yet we're assigned to do this!)
Thanks a bunch.
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Through minutes of toil, I got 2.40 grams of Aluminum Chloride needed. My next question is how much of the Sodium hydroxide and Ammonium hydroxide I need. They only tell me 3 M of a particular solution, so I'm a little confused as to how to get an amount.
3 M meaning molarity. Molarity= Moles of solute/Liters of solution
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Write a balanced chemical equation(s).
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Through minutes of toil, I got 2.40 grams of Aluminum Chloride needed. My next question is how much of the Sodium hydroxide and Ammonium hydroxide I need. They only tell me 3 M of a particular solution, so I'm a little confused as to how to get an amount.
3 M meaning molarity. Molarity= Moles of solute/Liters of solution
Yes, Molarity is moles of solute/liter or solution. When I calculate the mass of aluminum chloride needed for 5mL of a 3M solution, I get a different number. Can you post your calculations so I can see what went wrong?
To calculate the masses of sodium hydroxide and ammonium hydroxide you need you would first need to specify the volume of solution you want.
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You have to prepare solutions by youirself, or you will be given ready 3M solutions to use?
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Yes, Molarity is moles of solute/liter or solution. When I calculate the mass of aluminum chloride needed for 5mL of a 3M solution, I get a different number. Can you post your calculations so I can see what went wrong?
To calculate the masses of sodium hydroxide and ammonium hydroxide you need you would first need to specify the volume of solution you want.
I basically fiddled with the numbers a little to come up with an answer... any answer. I needed to feel that I've done something for the day. Haha. Please excuse silly, juvenile mistakes.
5 mL= .005 L
.005 L x 3 moles/1 L = .015 moles of Aluminum Chloride (160.32208 g Aluminum Chloride/ 1 Mole of Aluminum Chloride)= 2.40 g Aluminum Chloride?
I think I will be given the solutions ready to use, but I just have to specify how much of each I need. We will basically submit an order form of the chemicals that are required to execute the lab. Included in the "order form" are the solutions and how much of each will be needed and the lab equipment which isn't a problem. Also, the Aluminum Chloride will be in a solid form, I believe, while the sodium hydroxide and ammonium hydroxide solutions will be liquid.
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MM of aluminium chloride is about 132,5
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oh goodness, you're right. I had Al2Cl3 put down. Silly me.
The mass I calculated from Aluminum Chloride was 133.34 g/ mol.
That would mean that the calculations would change to 2.00 grams of Aluminum Chloride needed?
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For one 5 mL sample - you have 2
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Well, the amounted needed turned out to be 2.00 grams of aluminum chloride. Since I need to put that amount in TWO test tubes, I requested 4.00 grams. The 3M of Sodium Hydroxide and 3 M of Ammonium Hydroxide were requested in 2.5 mL each. If anyone sees error in this, please let me know! Thanks ;D
The question remains... why did one case precipitate and not in the other? ???
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This is an ibntrinsic property of Al(OH)3. It dissolves in an excess of OH- ions. NH3 is unable to form such a concetration that dissolves a significant amount of Al(OH)3
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Thanks, guys. I'll have to do some more research to explain the results in greater depth... and if anyone can add to AWK's answer, that would be fantastic.
I don't really know how to dilute solutions. If my teacher only prepares the sodium hydroxide and ammonium hydroxide in 6M and I need it in 3M, how much water do I need to add to each of those solutions?
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Apparently going from 6M to 3M is just diluting it by half. Since I requested 2.5 mL of each solution, I will just need to add 2.5 mL of water to the 2.5 mL in each solution to get the desired 3M solution. Is this right?
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Apparently going from 6M to 3M is just diluting it by half. Since I requested 2.5 mL of each solution, I will just need to add 2.5 mL of water to the 2.5 mL in each solution to get the desired 3M solution. Is this right?
Precisely!
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Apparently going from 6M to 3M is just diluting it by half. Since I requested 2.5 mL of each solution, I will just need to add 2.5 mL of water to the 2.5 mL in each solution to get the desired 3M solution. Is this right?
Completely off ;)
You are assuming that solution volumes are additive and that final volume will be 5 mL. At this accuracy level you are right, but in general solution volumes are NOT additive. The more concentrated solution, the larger the difference.
See attached screenshot of concentration calculator that takes these things into account. If you mix 2.500 mL of water and 2.500 mL of 6M hydrochloric acid final volume is not 5.000 but 4.992 mL, thus final concentration is not 3.000 but 3.005 M. Difference is negligibile and you may safely ignore it - as long as you remember it exists ;)
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In general, Borek is absolutely right, but 0.16 % error in volume or 0.17 % error in concentration are commonly accepted.
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Ahh.. I see. I carried out the experiment last Friday and basically got the desired results. The NaOH dissolved while the NH4OH did not. Could someone please elaborate on what AWK responded with on why one dissolved and one didn't? I looked up intrinsic properties and did not get very far.
What I was told was:
The reason that the precipitate dissolves in one of the solutions is because the equilibrium moves to the right until Al(OH)3(s) is precipitated; this easily reacts with excess NaOH to give the aluminate ion, Al(OH)4-(aq). Basically, the Al(OH)3 is complexed to create an ion that dissolves in solution.
I do not know if this answer is credible, but someone said that was the reason.
The chemical equation for this is a double replacement for both solutions, is it not?
*flummoxed* ???
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Yes, for both, then formation of [Al(OH)]4- complex with NaOH